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Physics Part-II

Nuclei

We are given the following atomic masses: { }_{92}^{238} {U}=2

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

Tritium has a half-life of

12.5

y undergoing beta decay. What fraction of a sample of pure tritium will remain undecayed after

25 y

We are given the following atomic masses:

_{92}^{238} U = 238.05079 u_{2}^{4} He = 4.00260 u

_{90}^{234} Th = 234.04363 u_{1}^{1} H = 1.00783 u

_{91}^{237} Pa = 237.05121 u

Here the symbol Pa is for the element protactinium

(Z = 91)

. (a) Calculate the energy released during the alpha decay of

_{92}^{238} U

. (b) Show that

_{92}^{238} U

can not spontaneously emit a proton.

Answer the following questions: (a) Are the equations of nuclear reactions (such as those given in Section 13.7) 'balanced' in the sense a chemical equation (e.g.,

2 H_{2} + O_{2} \to 2 H_{2} O

) is? If not, in what sense are they balanced on both sides? (b) If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction? (c) A general impression exists that mass-energy interconversion takes place only in nuclear reaction and never in chemical reaction. This is strictly speaking, incorrect. Explain.

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

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Solving time: 3 mins

We are given the following atomic masses: $_{92}^{238} U = 238.05079 u_{2}^{4} He = 4.00260 u$ $_{90}^{234} Th = 234.04363 u_{1}^{1} H = 1.00783 u$ $_{91}^{237} Pa = 237.05121 u$ Here the symbol Pa is for the element protactinium $(Z = 91)$ . (a) Calculate the energy released during the alpha decay of $_{92}^{238} U$ . (b) Show that $_{92}^{238} U$ can not spontaneously emit a proton.

Name: Video solution 1: [Solved] We are given the following atomic masses: 92238U=238.05079u24He=4.00260u 90234Th=234.04363u11H=1.00783u 91237 Pa=237.05121u Here the symbol Pa is for the element protactinium (Z=91). (a) Calculate the energy released during the alpha decay of 92238U. (b) Show that 92238U can not spontaneously emit a proton. - Physics Part-II
Uploaded: 2022-05-15T09:08:41.000Z
Duration: 8 min
Description: Video solution 1: [Solved] We are given the following atomic masses: 92238U=238.05079u24He=4.00260u 90234Th=234.04363u11H=1.00783u 91237 Pa=237.05121u Here the symbol Pa is for the element protactinium (Z=91). (a) Calculate the energy released during the alpha decay of 92238U. (b) Show that 92238U can not spontaneously emit a proton. - Physics Part-II

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Updated on: May 15, 2022

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Physics Part-II (NCERT)

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We are given the following atomic masses:

_{92}^{238} U = 238.05079 u_{2}^{4} He = 4.00260 u

_{90}^{234} Th = 234.04363 u_{1}^{1} H = 1.00783 u

_{91}^{237} Pa = 237.05121 u

Here the symbol Pa is for the element protactinium

(Z = 91)

. (a) Calculate the energy released during the alpha decay of

_{92}^{238} U

. (b) Show that

_{92}^{238} U

can not spontaneously emit a proton.

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

(a) The alpha decay of

_{92}^{238} U

is given by Eq. (13.20). The energy released in this process is given by

Q = (M_{U} - M_{Th} - M_{He}) c^{2}

Substituting the atomic masses as given in the data, we find

Q = (238.05079 - 234.04363 - 4.00260) u \times c^{2}

= (0.00456 u) c^{2}

= (0.00456 u) (931.5 MeV / u)

= 4.25 MeV .

(b) If

_{92}^{238} U

spontaneously emits a proton, the decay process would be

_{92}^{238} U \to_{91}^{237} Pa +_{1}^{1} H

The

Q

for this process to happen is

= (M_{U} - M_{Pa} - M_{H}) c^{2}

= (238.05079 - 237.05121 - 1.00783) u \times c^{2}

= (- 0.00825 u) c^{2}

= - (0.00825 u) (931.5 MeV / u)

= - 7.68 MeV

Thus, the

Q

of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of

7.68 MeV

to a

_{92}^{238} U

nucleus to make it emit a proton.

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Practice more questions from Physics Part-II (NCERT)

Tritium has a half-life of

12.5

y undergoing beta decay. What fraction of a sample of pure tritium will remain undecayed after

25 y

We are given the following atomic masses:

_{92}^{238} U = 238.05079 u_{2}^{4} He = 4.00260 u

_{90}^{234} Th = 234.04363 u_{1}^{1} H = 1.00783 u

_{91}^{237} Pa = 237.05121 u

Here the symbol Pa is for the element protactinium

(Z = 91)

. (a) Calculate the energy released during the alpha decay of

_{92}^{238} U

. (b) Show that

_{92}^{238} U

can not spontaneously emit a proton.

Answer the following questions: (a) Are the equations of nuclear reactions (such as those given in Section 13.7) 'balanced' in the sense a chemical equation (e.g.,

2 H_{2} + O_{2} \to 2 H_{2} O

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Question Text	We are given the following atomic masses: $_{92}^{238} U = 238.05079 u_{2}^{4} He = 4.00260 u$ $_{90}^{234} Th = 234.04363 u_{1}^{1} H = 1.00783 u$ $_{91}^{237} Pa = 237.05121 u$ Here the symbol Pa is for the element protactinium $(Z = 91)$ . (a) Calculate the energy released during the alpha decay of $_{92}^{238} U$ . (b) Show that $_{92}^{238} U$ can not spontaneously emit a proton.
Updated On	May 15, 2022
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	137
Avg. Video Duration	8 min

Text solutionVerified

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