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Page No 465: - Chapter 13 Nuclei Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024]


Question 13.27:

Consider the fission of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_575c4984.gif by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_6dc0cdec.gif  andhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_79618776.gif . Calculate Q for this fission process. The relevant atomic and particle masses are

mhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_30330c05.gif  =238.05079 u

mhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m22ff0521.gif  =139.90543 u

mhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_3b0e1df8.gif  = 98.90594 u

Answer:

In the fission ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_575c4984.gif , 10 β− particles decay from the parent nucleus. The nuclear reaction can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m762eee46.gif

It is given that:

Mass of a nucleushttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_49cafbc1.gif m1 = 238.05079 u

Mass of a nucleus https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m590c141e.gif m2 = 139.90543 u

Mass of a nucleushttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_79618776.gifm3 = 98.90594 u

Mass of a neutronhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m784a750a.gif m4 = 1.008665 u

Q-value of the above equation,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m94de8cf.gif

Where,

m’ = Represents the corresponding atomic masses of the nuclei

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1155cbfd.gifm1 − 92me

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m396cd546.gifm2 − 58me

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m60a6be48.gifm3 − 44me

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_783622de.gif m4

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8027/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_77fb9ffe.gif

Hence, the Q-value of the fission process is 231.007 MeV.

Question 13.28:

Consider the D−T reaction (deuterium−tritium fusion)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4475d22f.gif

(a) Calculate the energy released in MeV in this reaction from the data:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m2740f059.gif = 2.014102 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m44f710a1.gif = 3.016049 u

(b)Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); = Boltzman’s constant, = absolute temperature.)

Answer:

(a) Take the D-T nuclear reaction: https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_310752a1.gif

It is given that:

Mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_3680d9fb.gifm1= 2.014102 u

Mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m5aefea4d.gif , m= 3.016049 u

Mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7cba33cf.gif  m= 4.002603 u

Mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m81a25df.gifm= 1.008665 u

Q-value of the given D-T reaction is:

Q = [mm2− m3 − m4c2

= [2.014102 + 3.016049 − 4.002603 − 1.008665] c2

= [0.018883 c2] u

But 1 u = 931.5 MeV/c2

Q = 0.018883 × 931.5 = 17.59 MeV

(b) Radius of deuterium and tritium, r ≈ 2.0 fm = 2 × 10−15 m

Distance between the two nuclei at the moment when they touch each other, d = r + r = 4 × 10−15 m

Charge on the deuterium nucleus = e

Charge on the tritium nucleus = e

Hence, the repulsive potential energy between the two nuclei is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_6c4972b7.gif

Where,

0 = Permittivity of free space

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_157dcad2.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m50089b76.gif

Hence, 5.76 × 10−14 J or https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m5b5aacf3.gif of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.

However, it is given that:

KEhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1d7181a4.gif

Where,

k = Boltzmann constant = 1.38 × 10−23 m2 kg s−2 K−1

T = Temperature required for triggering the reaction

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8030/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m6b2326c9.gif

Hence, the gas must be heated to a temperature of 1.39 × 109 K to initiate the reaction.

Question 13.29:

Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that

(198Au) = 197.968233 u

(198Hg) =197.966760 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8032/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m30540cef.jpg

Answer:

It can be observed from the given γ-decay diagram that γ1 decays from the 1.088 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ1-decay is given as:

E1 = 1.088 − 0 = 1.088 MeV

1= 1.088 × 1.6 × 10−19 × 106 J

Where,

h = Planck’s constant = 6.6 × 10−34 Js

ν= Frequency of radiation radiated by γ1-decay

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8032/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_19d21a85.gif

It can be observed from the given γ-decay diagram that γ2 decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ2-decay is given as:

E2 = 0.412 − 0 = 0.412 MeV

2= 0.412 × 1.6 × 10−19 × 106 J

Where,

ν= Frequency of radiation radiated by γ2-decay

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8032/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7284339c.gif

It can be observed from the given γ-decay diagram that γ3 decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

Hence, the energy corresponding to γ3-decay is given as:

E3 = 1.088 − 0.412 = 0.676 MeV

3= 0.676 × 10−19 × 106 J

Where,

ν= Frequency of radiation radiated by γ3-decay

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8032/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m38c1cbd.gif

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8032/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_65b9a983.gif = 197.968233 u

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8032/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m2c7e389c.gif = 197.966760 u

1 u = 931.5 MeV/c2

Energy of the highest level is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8032/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m517c074a.gif

βdecays from the 1.3720995 MeV level to the 1.088 MeV level

∴Maximum kinetic energy of the βparticle = 1.3720995 − 1.088

= 0.2840995 MeV

βdecays from the 1.3720995 MeV level to the 0.412 MeV level

∴Maximum kinetic energy of the βparticle = 1.3720995 − 0.412

= 0.9600995 MeV

Also Read : Page-No-466:-Chapter-13-Nuclei-Additional-Exercise-Solutions-class-12-ncert-solutions-Physics

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