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INTRODUCTION - Chapter 13 Nuclei Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024]


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.gif have 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.gif andhttps://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.gif is 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.gif and

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.gif have 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.gif in 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 = Δmc2

= 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.gif atoms (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.gif atoms 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.gif nucleus 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.gif and 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.gif dating 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.

Also Read : Page-No-463:-Chapter-13-Nuclei-Exercise-Solutions-class-12-ncert-solutions-Physics

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