13- NUCLEI


Here we are going through the ncert solutions for class 12 physics chapter 13- nuclei so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ Chapter 13 - Nuclei

 

Question 13.1:

(a) Two stable isotopes of lithium https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_mc5f3b0a.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6ed52b7e.gifhave respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Boron has two stable isotopes, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1811c7c1.gifandhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m284928f1.gif. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1811c7c1.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m284928f1.gif.

ANSWER:

(a) Mass of lithium isotope https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_mc5f3b0a.gifm1 = 6.01512 u

Mass of lithium isotope https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6ed52b7e.gifm2 = 7.01600 u

Abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_mc5f3b0a.gifη1= 7.5%

Abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6ed52b7e.gifη2= 92.5%

The atomic mass of lithium atom is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7a9ef414.gif

(b) Mass of boron isotope https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1811c7c1.gifm1 = 10.01294 u

Mass of boron isotope https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m284928f1.gifm2 = 11.00931 u

Abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1811c7c1.gifη1 = x%

Abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m284928f1.gifη2= (100 − x)%

Atomic mass of boron, m = 10.811 u

The atomic mass of boron atom is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1de7a095.gif

And 100 − x = 80.11%

Hence, the abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1811c7c1.gif is 19.89% and that of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7976/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m284928f1.gifis 80.11%.

 

Question 13.2:

The three stable isotopes of neon: https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m67c5f9b0.gifand https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7516ccb8.gifhave respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

ANSWER:

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m731a6c7c.gifm1= 19.99 u

Abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m731a6c7c.gifη= 90.51%

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m41af246.gif, m= 20.99 u

Abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m41af246.gifη= 0.27%

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7516ccb8.gifm= 21.99 u

Abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7516ccb8.gifη3 = 9.22%

The average atomic mass of neon is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7983/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m42b4bf1d.gif

 

Question 13.3:

Obtain the binding energy (in MeV) of a nitrogen nucleushttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7984/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_69ad3574.gif, given https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7984/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_a3f78c2.gif=14.00307 u

ANSWER:

Atomic mass of nitrogenhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7984/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_48b167c.gifm = 14.00307 u

A nucleus of nitrogen https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7984/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m68d62598.gif contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δ= 7mH + 7mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn= 1.008665 u

Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But 1 u = 931.5 MeV/c2

Δ= 0.11236 × 931.5 MeV/c2

Hence, the binding energy of the nucleus is given as:

Eb = Δmc2

Where,

c = Speed of light

E= 0.11236 × 931.5 https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7984/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m47760016.gif

= 104.66334 MeV

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

 

Question 13.4:

Obtain the binding energy of the nuclei https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7468777b.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m515abbc8.gifin units of MeV from the following data:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m53f6f187.gif = 55.934939 u https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_75e517fd.gif= 208.980388 u

ANSWER:

Atomic mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7468777b.gifm1 = 55.934939 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7468777b.gif nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, Δ= 26 × mH + 30 × mn − m1

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

Δ= 26 × 1.007825 + 30 × 1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV/c2

Δ= 0.528461 × 931.5 MeV/c2

The binding energy of this nucleus is given as:

Eb1 = Δmc2

Where,

c = Speed of light

Eb1 = 0.528461 × 931.5 https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m47760016.gif

= 492.26 MeV

Average binding energy per nucleon https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6b5d739f.gif

Atomic mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m515abbc8.gifm2 = 208.980388 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m515abbc8.gif nucleus has 83 protons and (209 − 83) 126 neutrons.

Hence, the mass defect of this nucleus is given as:

Δm' = 83 × mH + 126 × mn − m2

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

Δm' = 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/c2

Δm' = 1.760877 × 931.5 MeV/c2

Hence, the binding energy of this nucleus is given as:

Eb2 = Δm'c2

= 1.760877 × 931.5https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m47760016.gif

= 1640.26 MeV

Average bindingenergy per nucleon = https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7985/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m25a5875d.gif

 

Question 13.5:

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7986/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_12315718.gifatoms (of mass 62.92960 u).

ANSWER:

Mass of a copper coin, m’ = 3 g

Atomic mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7986/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m74fcda2d.gif atom, m = 62.92960 u

The total number of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7986/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m74fcda2d.gifatoms in the coinhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7986/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m44478d7e.gif

Where,

NA = Avogadro’s number = 6.023 × 1023 atoms /g

Mass number = 63 g

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7986/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m63c073bc.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7986/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m74fcda2d.gifnucleus has 29 protons and (63 − 29) 34 neutrons

Mass defect of this nucleus, Δm' = 29 × mH + 34 × mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

Δm' = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022

= 1.69766958 × 1022 u

But 1 u = 931.5 MeV/c2

Δ= 1.69766958 × 1022 × 931.5 MeV/c2

Hence, the binding energy of the nuclei of the coin is given as:

Eb= Δmc2

= 1.69766958 × 1022 × 931.5 https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7986/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m47760016.gif

= 1.581 × 1025 MeV

But 1 MeV = 1.6 × 10−13 J

Eb = 1.581 × 1025 × 1.6 × 10−13

= 2.5296 × 1012 J

This much energy is required to separate all the neutrons and protons from the given coin.

 

Question 13.6:

Write nuclear reaction equations for

(i) α-decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_macfc9e9.gif (ii) α-decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4cc04352.gif

(iii) β-decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5db0e64a.gif (iv) β-decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m135441be.gif

(v) β+-decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2df1b845.gif (vi) β+-decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m45a87636.gif

(vii) Electron capture of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6259ecd4.gif

ANSWER:

α is a nucleus of helium https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1b89935a.gifand β is an electron (e− for β and e+ for β+). In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

For the given cases, the various nuclear reactions can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7988/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m3bddd8d1.gif

 

Question 13.7:

A radioactive isotope has a half-life of years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

ANSWER:

Half-life of the radioactive isotope = T years

Original amount of the radioactive isotope = N0

(a) After decay, the amount of the radioactive isotope = N

It is given that only 3.125% of Nremains after decay. Hence, we can write:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7989/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_871f12c.gif

Where,

λ = Decay constant

t = Time

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7989/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_37403eec.gif

Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.

(b) After decay, the amount of the radioactive isotope = N

It is given that only 1% of Nremains after decay. Hence, we can write:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7989/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5429cd5b.gif

Since, λ = 0.693/T

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7989/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_c5ea9d.gif

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

 

Question 13.8:

The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7991/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_10871d78.gif present with the stable carbon isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7991/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_308c5d93.gif . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7991/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_10871d78.gif, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7991/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_10871d78.gifdating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

ANSWER:

Decay rate of living carbon-containing matter, R = 15 decay/min

Let N be the number of radioactive atoms present in a normal carbon- containing matter.

Half life ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7991/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_10871d78.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7991/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_mdda5227.gif= 5730 years

The decay rate of the specimen obtained from the Mohenjodaro site:

R' = 9 decays/min

Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro period.

Therefore, we can relate the decay constant, λand time, t as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7991/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m17fa739b.gif

Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.


Question 13.9:

Obtain the amount of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m3f06cd34.gifnecessary to provide a radioactive source of 8.0 mCi strength. The half-life of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m3f06cd34.gifis 5.3 years.

ANSWER:

The strength of the radioactive source is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m267cc491.gif

Where,

N = Required number of atoms

Half-life ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m3f06cd34.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_mdda5227.gif = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 108 s

For decay constant λ, we have the rate of decay as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6a4a579b.gif

Where, λ https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m570fa75d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4762210d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m5663814f.gif

Forhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_20db6e7b.gif:

Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m73d7ff59.gifatoms https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1d8aa47.gif

Hence, the amount of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7993/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_20db6e7b.gif necessary for the purpose is 7.106 × 10−6 g.

 

Question 13.10:

The half-life of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_679d9b6d.gifis 28 years. What is the disintegration rate of 15 mg of this isotope?

ANSWER:

Half life of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_679d9b6d.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m55c40a46.gif= 28 years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 108 s

Mass of the isotope, m = 15 mg

90 g of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_679d9b6d.gifatom contains 6.023 × 1023 (Avogadro’s number) atoms.

Therefore, 15 mg of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_679d9b6d.gif contains:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m63042820.gif

Rate of disintegration, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6a4a579b.gif

Where,

λ = Decay constant https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_bc61521.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7994/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_503e32e6.gif

Hence, the disintegration rate of 15 mg of the given isotope is
7.878 × 10
10 atoms/s.

 

Question 13.11:

Obtain approximately the ratio of the nuclear radii of the gold isotope https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7996/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_54d7899.gif and the silver isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7996/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_6d3d8473.gif.

ANSWER:

Nuclear radius of the gold isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7996/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2775a1f.gif = RAu

Nuclear radius of the silver isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7996/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_451673c1.gif = RAg

Mass number of gold, AAu = 197

Mass number of silver, AAg = 107

The ratio of the radii of the two nuclei is related with their mass numbers as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7996/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m5270376f.gif

Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

 

Question 13.12:

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4a5f44ab.gifand (b)https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4599dedc.gif.

Given https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_41aeaf34.gif = 226.02540 u, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_570b5a6b.gif = 222.01750 u,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_177ca0ef.gif= 220.01137 u, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_594538cc.gif= 216.00189 u.

ANSWER:

(a) Alpha particle decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m2ecc7672.gifemits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_27853c1f.gif

Q-value of

emitted α-particle = (Sum of initial mass − Sum of final mass) c2

Where,

c = Speed of light

It is given that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_403feaa6.gif

Q-value = [226.02540 − (222.01750 + 4.002603)] u c2
= 0.005297 u 
c2

But 1 u = 931.5 MeV/c2

Q = 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the α-particle https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m38e65129.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m2a706cc2.gif

(b) Alpha particle decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_52d61109.gif is shown by the following nuclear reaction.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_34726468.gif

It is given that:

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_52d61109.gif= 220.01137 u

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bd29de8.gif= 216.00189 u

Q-value = https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m70e1011.gif

≈ 641 MeV

Kinetic energy of the α-particle https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/7998/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m44397a56.gif

= 6.29 MeV

 

Question 13.13:

The radionuclide 11C decays according to

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_46d2dae0.gif

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m14726060.gif

calculate and compare it with the maximum energy of the positron emitted

ANSWER:

The given nuclear reaction is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m107aae15.gif

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m56a7e5bd.gif= 11.011434 u

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_51ec3683.gif

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m26f5a503.gif nucleus is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_72bdba2d.gif

Where,

me = Mass of an electron or positron = 0.000548 u

= Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_18f7c89a.gifand 5 min the case ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m74598a6e.gif.

Hence, equation (1) reduces to:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8000/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2887962a.gif

ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c2

= (0.001033 c2) u

But 1 u = 931.5 Mev/c2

ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron.

 

Question 13.14:

The nucleus https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_34ed9aa7.gifdecays byhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m67556fa3.gifemission. Write down thehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1de0ccf7.gif decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2d5bc804.gif= 22.994466 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6e3ecb1f.gif= 22.989770 u.

ANSWER:

In https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m67556fa3.gif emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m67556fa3.gif emission of the nucleus https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_34ed9aa7.gif is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m10ad2d46.gif

It is given that:

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2d5bc804.gif= 22.994466 u

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6e3ecb1f.gif= 22.989770 u

Mass of an electron, m= 0.000548 u

Q-value of the given reaction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_157ebe0f.gif

There are 10 electrons in https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_34ed9aa7.gif and 11 electrons inhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6358b7e7.gif. Hence, the mass of the electron is cancelled in the Q-value equation.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_63b20059.gif

The daughter nucleus is too heavy as compared to https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_6b7ff0ec.gifand https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8002/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1403ccce.gif. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.

 

Question 13.15:

The value of a nuclear reaction → is defined by

= [ mAmb− mC− md]cwhere the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_616a61bf.gif

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1b33adcf.gif

Atomic masses are given to be

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4705f282.gif

ANSWER:

(i) The given nuclear reaction is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_63c8145.gif

It is given that:

Atomic mass https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_6f8a4e56.gif

Atomic mass https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7f7ffd92.gif

Atomic mass https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_3c4e1166.gif

According to the question, the Q-value of the reaction can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5174426f.gif

The negativeQ-value of the reaction shows that the reaction is endothermic.

(ii) The given nuclear reaction is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_3625d78e.gif

It is given that:

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m2561efac.gif

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_623aeb77.gif

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m57c1b040.gif

The Q-value of this reaction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8005/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m55d49e49.gif

The positive Q-value of the reaction shows that the reaction is exothermic.

 

Question 13.16:

Suppose, we think of fission of a https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4037bd97.gifnucleus into two equal fragments,https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m22a24db5.gif. Is the fission energetically possible? Argue by working out of the process. Given https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1013b8fe.gif andhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_289f5e87.gif.

ANSWER:

The fission of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4037bd97.gifcan be given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_13ceb8b2.gif

It is given that:

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m2e0e01c1.gif = 55.93494 u

Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m128810ac.gif

The Q-value of this nuclear reaction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8007/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6409a453.gif

The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.

 

Question 13.17:

The fission properties of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m36efaa7f.gifare very similar to those ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gif.

The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m36efaa7f.gifundergo fission?

ANSWER:

Average energy released per fission ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m36efaa7f.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1e6c7ed8.gif

Amount of purehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_me69179b.gif, m = 1 kg = 1000 g

NA= Avogadro number = 6.023 × 1023

Mass number ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m36efaa7f.gif= 239 g

1 mole of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_me69179b.gifcontains NA atoms.

m g of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_me69179b.gifcontainshttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6bec6381.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m66c0a3ef.gif

Total energy released during the fission of 1 kg ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m36efaa7f.gifis calculated as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4b87fbfd.gif

Hence, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2b0ab274.gif is released if all the atoms in 1 kg of pure https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8009/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_me69179b.gifundergo fission.



Question 13.18:

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gifdid it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gifand that this nuclide is consumed only by the fission process.

ANSWER:

Half life of the fuel of the fission reactor, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m17b65a4a.gifyears

= 5 × 365 × 24 × 60 × 60 s

We know that in the fission of 1 g of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gifnucleus, the energy released is equal to 200 MeV.

1 mole, i.e., 235 g of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gifcontains 6.023 × 1023 atoms.

1 g https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gif containshttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m265e86aa.gif

The total energy generated per gram ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gifis calculated as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_421c672c.gif

The reactor operates only 80% of the time.

Hence, the amount of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gifconsumed in 5 years by the 1000 MW fission reactor is calculated as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m3b19010a.gif

Initial amount of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8010/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4c809fc5.gif= 2 × 1538 = 3076 kg

 

Question 13.19:

How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8012/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1e1c12f7.gif

ANSWER:

The given fusion reaction is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8012/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m27c52578.gif

Amount of deuterium, m = 2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 × 1023 atoms.

2.0 kg of deuterium contains https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8012/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m5b3cdbf2.gif

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

Total energy per nucleus released in the fusion reaction:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8012/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_338b2ae8.gif

Power of the electric lamp, P = 100 W = 100 J/s

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8012/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2253470a.gif

 

Question 13.20:

Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

ANSWER:

When two deuterons collide head-on, the distance between their centres, d is given as:

Radius of 1st deuteron + Radius of 2nd deuteron

Radius of a deuteron nucleus = 2 fm = 2 × 10−15 m

d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m

Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C

Potential energy of the two-deuteron system:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8014/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_53cac0b9.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8014/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_149b60d9.gif = Permittivity of free space

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8014/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4a759303.gif

Hence, the height of the potential barrier of the two-deuteron system is

360 keV.

 

Question 13.21:

From the relation R0A1/3, where R0 is a constant and is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

ANSWER:

We have the expression for nuclear radius as:

R0A1/3

Where,

R0 = Constant.

A = Mass number of the nucleus

Nuclear matter density, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8016/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_64aed25d.gif

Let m be the average mass of the nucleus.

Hence, mass of the nucleus = mA

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8016/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m3b9a5ea7.gif

Hence, the nuclear matter density is independent of A. It is nearly constant.

 

Question 13.22:

For the https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1907344.gif (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K−shell, is captured by the nucleus and a neutrino is emitted).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_9618240.gif

Show that if https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1907344.gif emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.

ANSWER:

Let the amount of energy released during the electron capture process be Q1. The nuclear reaction can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m267ead8e.gif

Let the amount of energy released during the positron capture process be Q2. The nuclear reaction can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2c3617e9.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7f39da4f.gif= Nuclear mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_3afdd1df.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1df30fec.gif= Nuclear mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m5e040f70.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m5fe2947e.gif= Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_3afdd1df.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4afafe0c.gif= Atomic mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m5e040f70.gif

me = Mass of an electron

= Speed of light

Q-value of the electron capture reaction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7a33daf3.gif

Q-value of the positron capture reaction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1a73d8fc.gif

It can be inferred that if Q2 > 0, then Q> 0; Also, if Q1> 0, it does not necessarily mean that Q2 > 0.

In other words, this means that ifhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8017/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m1907344.gifemission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically-allowed nuclear reaction.

 

Question 13.23:

In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_24294bfb.gif(23.98504u), https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5b69b0f6.gif(24.98584u) and https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2225eb37.gif(25.98259u). The natural abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_24294bfb.gifis 78.99% by mass. Calculate the abundances of other two isotopes.

ANSWER:

Average atomic mass of magnesium, m = 24.312 u

Mass of magnesium isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_24294bfb.gifm1 = 23.98504 u

Mass of magnesium isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5b69b0f6.gif, m= 24.98584 u

Mass of magnesium isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2225eb37.gif, m= 25.98259 u

Abundance ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_24294bfb.gifη1= 78.99%

Abundance ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5b69b0f6.gifηx%

Hence, abundance ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2225eb37.gifη= 100 − x − 78.99% = (21.01 − x)%

We have the relation for the average atomic mass as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5096dd47.gif

Hence, the abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5b69b0f6.gifis 9.3% and that of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2225eb37.gifis 11.71%.

 

Question 13.24:

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_f5668ef.gifand https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_466374fc.giffrom the following data:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4b3911fa.gif= 39.962591 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1ebe578c.gif) = 40.962278 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7f4da733.gif= 25.986895 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m727d36a5.gif) = 26.981541 u

ANSWER:

For https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m862cde1.gif

Forhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5dc60d5b.gif

A neutron https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m46e1a79a.gifis removed from ahttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_f5668ef.gif nucleus. The corresponding nuclear reaction can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m48efc39b.gif

It is given that:

Mass https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4b3911fa.gif= 39.962591 u

Masshttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1ebe578c.gif) = 40.962278 u

Mass