[Solved] Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:88223Ra→82209Pb+614C88223Ra→86219Rn+24HeCalculate the Q-values for these decays and detere that both are energetically allowed. - Physics Part-II

(i) In case of first process,88223Ra→ 20982Pb+614C+QThe mass defect occured in reaction is,Δm=mass of Ra223−(mass of Pb209+mass of C14)=223.01850−(208.98107+14.00324)=0.03419uSo, the amount of energy released is given by:Q=Δm×931 MeVQ=0.03419×931MeV=31.83MeV(ii) Now, in case of second reaction,88223Ra→ 86219Rn+24He+QMass defect is given as:Δm= mass of Ra223−(mass of Rn219+ mass of He4)=223.01850−(219.00948+4.00260)=0.00642u∴ the Energy released in the reaction is:Q=0.00642×931MeV=5.98MeVSince both reactions have positive Q-value so both reactions are possible.

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Physics Part-II

Nuclei

Under certain circumstances, a nucleus can decay by emitting a

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

A source contains two phosphorous radio nuclides

_{15}^{32} P (T_{1/2} = 14.3 d)

and

_{15}^{33} P (T_{1/2} = 25.3 d)

. Initially, 10% of the decays come from

_{15}^{32} P

. How long one must wait until 90% to do so?

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

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

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Solving time: 4 mins

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.

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Physics Part-II (NCERT)

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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.

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

(i) In case of first process,

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

The mass defect occured in reaction is,

Δ m = ma ss o f R a^{223} - (ma ss o f P b^{209} + ma ss o f C^{14})

= 223.01850 - (208.98107 + 14.00324)

= 0.03419 u

So, the amount of energy released is given by:

Q = Δ m \times 931 M e V

Q = 0.03419 \times 931 M e V = 31.83 M e V

(ii) Now, in case of second reaction,

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

Mass defect is given as:

Δ m = ma ss o f R a^{223} - (ma ss o f R n^{219} + ma ss o f H e^{4})

= 223.01850 - (219.00948 + 4.00260)

= 0.00642 u

∴

the Energy released in the reaction is:

Q = 0.00642 \times 931 M e V = 5.98 M e V

Since both reactions have positive Q-value so both reactions are possible.

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Practice more questions from Physics Part-II (NCERT)

A source contains two phosphorous radio nuclides

_{15}^{32} P (T_{1/2} = 14.3 d)

and

_{15}^{33} P (T_{1/2} = 25.3 d)

. Initially, 10% of the decays come from

_{15}^{32} P

. How long one must wait until 90% to do so?

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

View all

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Question Text	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.
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	11

Text solutionVerified

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