Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron =1.6725×10−27 kg, mass of proton =1.6725×10−27 kg, mass of electron =9×10−31 kg)(a) 0.73MeV(b) 7.10MeV(c) 6.30MeV(d) 5.4MeV

Name: Video solution 1: Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron =1.6725×10−27 kg, mass of proton =1.6725×10−27 kg, mass of electron =9×10−31 kg)(a) 0.73MeV(b) 7.10MeV(c) 6.30MeV(d) 5.4MeV
Uploaded: 2023-01-05T20:25:38.000Z
Duration: 20 min
Description: Video solution 1: Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron =1.6725×10−27 kg, mass of proton =1.6725×10−27 kg, mass of electron =9×10−31 kg)(a) 0.73MeV(b) 7.10MeV(c) 6.30MeV(d) 5.4MeV

Exp. (a)According to given data, mass of neutron and proton are equal which do not permit the breaking up of neutron and proton.But if we take standard mass of neutron as 1.6750×10−27 kg, thenEnergy released = Mass defect ×c2=(mn−mp−me)×c2[∵ energy released mass of reactant − mass of product ×C2 ]=1.66×10−27(1.6750×10−27−1.6725×10−27−9×10−31)×931.5MeV[∵1amu=931.5MeV]=1.66×10−2716×10−31×931.5MeV=1.661.6×0.9315MeV≈0.9MeVwhich is close to option (a).

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Mass-Energy and Nuclear Binding Energy

Assume that a neutron breaks into a proton and an electron. Th

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Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron $= 1.6725 \times 1 0^{- 27} kg$ , mass of proton $= 1.6725 \times 1 0^{- 27} kg$ , mass of electron $= 9 \times 1 0^{- 31} kg)$
(a) $0.73 MeV$
(b) $7.10 MeV$
(c) $6.30 MeV$
(d) $5.4 MeV$

Views: 6,159 students

Updated on: Jan 5, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2012

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Benjamin Discussed

Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron

= 1.6725 \times 1 0^{- 27} kg

, mass of proton

= 1.6725 \times 1 0^{- 27} kg

, mass of electron

= 9 \times 1 0^{- 31} kg)

(a)

0.73 MeV

(b)

7.10 MeV

(c)

6.30 MeV

(d)

5.4 MeV

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

Exp. (a)
According to given data, mass of neutron and proton are equal which do not permit the breaking up of neutron and proton.
But if we take standard mass of neutron as

1.6750 \times 1 0^{- 27} kg

, then
Energy released

=

Mass defect

\times c^{2}

= (m_{n} - m_{p} - m_{e}) \times c^{2}

[∵

energy released mass of reactant

-

mass of product

\times C^{2}

]

= \frac{( 1.6750 \times 1 0 ^{- 27} - 1.6725 \times 1 0 ^{- 27} - 9 \times 1 0 ^{- 31} )}{1.66 \times 1 0 ^{- 27}}

\times 931.5 MeV [∵ 1 amu = 931.5 MeV]

= \frac{16 \times 1 0 ^{- 31}}{1.66 \times 1 0 ^{- 27}} \times 931.5 MeV

= \frac{1.6 \times 0.9315}{1.66} MeV

\approx 0.9 MeV

which is close to option (a).

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Question Text	Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron $= 1.6725 \times 1 0^{- 27} kg$ , mass of proton $= 1.6725 \times 1 0^{- 27} kg$ , mass of electron $= 9 \times 1 0^{- 31} kg)$ (a) $0.73 MeV$ (b) $7.10 MeV$ (c) $6.30 MeV$ (d) $5.4 MeV$
Updated On	Jan 5, 2023
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	141
Avg. Video Duration	20 min

Text solutionVerified

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Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron =1.6725×10−27 kg, mass of proton =1.6725×10−27 kg, mass of electron =9×10−31 kg)(a) 0.73MeV(b) 7.10MeV(c) 6.30MeV(d) 5.4MeV

Text solutionVerified

Filo tutor solutions (1)

Questions from JEE Mains 2012 - PYQs

Practice questions from Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

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