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Physics Part-II

Nuclei

Obtain the maximum kinetic energy of -particles, and the radi

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

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.)

Obtain the maximum kinetic energy of

β

-particles, and the radiation frequencies of

γ

decays in the decay scheme shown.

m (^{198} A u) = 197.968233 u

Calculate and compare the energy released by (a) Fusion of 1.0 kg of hydrogen deep within Sun (b) The fission of 1.0 kg of

^{235} U

in a fission reactor

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

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Solving time: 6 mins

Obtain the maximum kinetic energy of $β$ -particles, and the radiation frequencies of $γ$ decays in the decay scheme shown. $m (^{198} A u) = 197.968233 u$

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Physics Part-II (NCERT)

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Obtain the maximum kinetic energy of

β

-particles, and the radiation frequencies of

γ

decays in the decay scheme shown.

m (^{198} A u) = 197.968233 u

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

It can be observed from the given

γ

decay diagram that

γ_{1}

decays from the 1.088 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to

γ_{1}

decays is given as

E_{1} = 1.088 - 0 = 1.088 M e V

h v_{1} = 1.088 \times 1.6 \times 1 0^{- 19} \times 1 0^{6} J

Where,

h= Plank's constant and

v_{1}

= frequency of radiation radiated by

γ_{1}

decay.

∴ v_{1} = \frac{E _{1}}{h} = \frac{1.088 \times 1.6 \times 1 0 ^{19} \times 1 0 ^{6}}{6.6 \times 1 0 ^{- 34}} = 2.637 \times 1 0^{20} Hz

It can be observed the given gamma decay diagram that

γ_{2}

decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to

γ_{2}

decay is given as

E_{2} = 0.412 - 0 = 0.412 M e V

h v_{2} = 0.412 \times 1.6 \times 1 0^{- 19} \times 1 0^{6} J

Where

v_{2}

is the frequency of the radiation radiated by

γ_{2}

decay.

∴ v_{2} = \frac{E _{2}}{h} = 9.988 \times 1 0^{19} Hz

It can be observed from the given gamma decay diagram that

γ_{3}

decays from the 1.088 MeV energy level to the 0.412 MeV energy level. Hence, the energy corresponding to

γ_{3}

is given as

E_{3} = 1.088 - 0.412 = m 0.676 M e V J

h v_{3} = 0.676 \times 1 0^{- 19} \times 1 0^{6} J

Where

v_{3} =

frequency of the radiation radiated by

γ_{3}

decay

∴ v_{3} = \frac{E _{3}}{h} = 1.639 \times 1 0^{20} Hz

Now mass of

A u

is 197.968 u and mass of

H g

= 197.9667 u

1 u = 931.5 M e V / c^{2}

Energy of the highest level is given as :

E = [m (_{78}^{198} A u) - m (_{80}^{190} H g)] = 197.968233 - 197.96676 = 0.001473 u = 0.001473 \times 931.5 = 1.3720995 M e V

β_{1}

decays from the 1.3720995 MeV level to the 1.088 MeV level

∴

maximum kinetic energy of the

β_{1}

particle

= 1.3720995 - 1.088 = 0.2840995 M e V

β_{2}

decays from the 1.3720995 MeV level to the 0.412 MeV level

∴

Maximum kinetic energy of the

β_{2}

particle =

1.3720995 - 0.412 = 0.9600995

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Practice more questions from Physics Part-II (NCERT)

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

Obtain the maximum kinetic energy of

β

-particles, and the radiation frequencies of

γ

decays in the decay scheme shown.

m (^{198} A u) = 197.968233 u

Calculate and compare the energy released by (a) Fusion of 1.0 kg of hydrogen deep within Sun (b) The fission of 1.0 kg of

^{235} U

in a fission reactor

View all

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Question Text	Obtain the maximum kinetic energy of $β$ -particles, and the radiation frequencies of $γ$ decays in the decay scheme shown. $m (^{198} A u) = 197.968233 u$
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	11

Obtain the maximum kinetic energy of $β$ -particles, and the radiation frequencies of $γ$ decays in the decay scheme shown. $m (^{198} A u) = 197.968233 u$

Text solutionVerified

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Obtain the maximum kinetic energy of $β$ -particles, and the radiation frequencies of $γ$ decays in the decay scheme shown. $m (^{198} A u) = 197.968233 u$