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Physics Part-II

Nuclei

Consider the fission of ^{238}_{92}U by fast neutrons. In one

Physics Part-II

Class 12

NCERT

Chapter 1: Ray Optics and Optical Instruments

41 questions

Chapter 2: Wave Optics

30 questions

Chapter 3: Dual Nature of Radiation and Matter

44 questions

Chapter 4: Atoms

23 questions

Chapter 5: Nuclei

38 questions

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an

α

-particle. Consider the following decay processes:

_{88}^{223} R a \to_{82}^{209} P b +_{6}^{14} C

_{88}^{223} R a \to_{86}^{219} R n +_{2}^{4} He

Calculate the Q-values for these decays and detere that both are energetically allowed.

Consider the fission of

_{92}^{238} U

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

_{58}^{140} C e

and

_{44}^{99} R u

. Calculate Q for this fission process. The relevant atomic and particle masses are:

m (_{92}^{238} U) = 238.05079 u

m (_{58}^{140} C e) = 139.90543 u

m (_{44}^{99} U) = 98.90594 u

Consider the DT reaction (deuteriumtritium fusion).

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He + n

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

m (_{1}^{2} H) = 2.014102 u

m (_{1}^{3} H) = 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); k = Boltzmans constant, T = absolute temperature.)

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Chapter 6: Semiconductor Electronics: Materials, Devices and Simple Circuits

25 questions

Question

Easy

Solving time: 4 mins

Consider the fission of $_{92}^{238} U$ 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 $_{58}^{140} C e$ and $_{44}^{99} R u$ . Calculate Q for this fission process. The relevant atomic and particle masses are:
$m (_{92}^{238} U) = 238.05079 u$
$m (_{58}^{140} C e) = 139.90543 u$
$m (_{44}^{99} U) = 98.90594 u$ .

Name: Video solution 1: [Solved] Consider the fission of 92238U 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 58140Ce and 4499Ru. Calculate Q for this fission process. The relevant atomic and particle masses are:m(92238U)=238.05079um(58140Ce)=139.90543um(4499U)=98.90594u. - Physics Part-II
Uploaded: 2022-05-09T08:08:06.000Z
Duration: 5 min
Description: Video solution 1: [Solved] Consider the fission of 92238U 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 58140Ce and 4499Ru. Calculate Q for this fission process. The relevant atomic and particle masses are:m(92238U)=238.05079um(58140Ce)=139.90543um(4499U)=98.90594u. - Physics Part-II

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Updated on: May 9, 2022

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Physics Part-II (NCERT)

Found 2 tutors discussing this question

Liam Discussed

Consider the fission of

_{92}^{238} U

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

_{58}^{140} C e

and

_{44}^{99} R u

. Calculate Q for this fission process. The relevant atomic and particle masses are:

m (_{92}^{238} U) = 238.05079 u

m (_{58}^{140} C e) = 139.90543 u

m (_{44}^{99} U) = 98.90594 u

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

The reaction for the fission of

_{93}^{238} U

by fast neutrons is as shown

_{93}^{238} U +_{0}^{1} n \to_{58}^{140} C e +_{44}^{99} U

Q = [m (_{92}^{238} U) + m_{N} - m (_{58}^{140} C e) - m (_{44}^{99} R u)] \times

931 .5 MeV

Q = [238.05079 + 1.00893 - 139.90543 - 98.90594] \times

931 .5 MeV

Q = 0.24835 \times

931 .5 MeV

Q = 231.1

MeV

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Practice more questions from Physics Part-II (NCERT)

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an

α

-particle. Consider the following decay processes:

_{88}^{223} R a \to_{82}^{209} P b +_{6}^{14} C

_{88}^{223} R a \to_{86}^{219} R n +_{2}^{4} He

Calculate the Q-values for these decays and detere that both are energetically allowed.

Consider the fission of

_{92}^{238} U

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

_{58}^{140} C e

and

_{44}^{99} R u

. Calculate Q for this fission process. The relevant atomic and particle masses are:

m (_{92}^{238} U) = 238.05079 u

m (_{58}^{140} C e) = 139.90543 u

m (_{44}^{99} U) = 98.90594 u

Consider the DT reaction (deuteriumtritium fusion).

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He + n

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

m (_{1}^{2} H) = 2.014102 u

m (_{1}^{3} H) = 3.016049 u

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Question Text	Consider the fission of $_{92}^{238} U$ 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 $_{58}^{140} C e$ and $_{44}^{99} R u$ . Calculate Q for this fission process. The relevant atomic and particle masses are: $m (_{92}^{238} U) = 238.05079 u$ $m (_{58}^{140} C e) = 139.90543 u$ $m (_{44}^{99} U) = 98.90594 u$ .
Updated On	May 9, 2022
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	140
Avg. Video Duration	5 min

Text solutionVerified

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