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Standard XII

Physics

Binding Energy

Assume that a...

Question

Assume that a neutron breaks into a proton and an electron. The energy released during this process is
(Mass of neutron $= 1.6725 \times 10^{- 27} k g$ , Mass of proton $= 1.6725 \times 10^{- 27} k g$ ,Mass of electron $= 9 \times 10^{- 31} k g$ )

0.50625

M e V

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7.10

M e V

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6.30

M e V

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5.40

M e V

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Solution

The correct option is A $0.50625$ $M e V$
$n \to p + e$
$Δ m = (m_{p} + m_{e}) - m_{n}$
$Δ m = - 9 \times 10^{- 31} k g$
Energy released $= 9 \times 10^{- 31} \times (3 \times 10^{8})^{2} J$
$= \frac{9 \times 10^{- 31} \times (3 \times 10^{8})^{2}}{1.6 \times 10^{- 13}}$
$= 0.50625 M e V$

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Similar questions

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

= 1.6725 \times 10^{- 27} k g,

mass od proton

= 1.6725 \times 10^{- 27} k g,

mass of electron

= 9 \times 10^{- 31} k g)

Q. A neutron breaks into a proton and an electron. Calculate the energy produced in this reaction is MeV. Mass of an electron

= 9 \times 10^{- 31} k g

, mass of proton

= 1.6725 \times 10^{- 27} k g

, mass of neutron

= 1.6747 \times 10^{- 27} k g

, and speed of light

= 3 \times 10^{8} m s^{- 1}

Q. The rest mass of a deuteron is equivalent to an energy of

1876

MeV, that of a proton to

939

MeV, and that of a neutron to

940

MeV. A deuteron may disintegrate to a proton and a neutron if it

Suppose an electron beam is used to probe the internal structure of a neutron. The required kinetic energy of the electron is (assume the diameter of a neutron to be of the order of $10^{- 15} m$ ) of the order of

Q. The binding energy per nucleon of

_{5} B^{10}

8.0

MeV and that of

_{5} B^{11}

7.5

MeV. The Energy required to remove a neutron from

_{5} B^{11}

is (mass of electron and proton are

9.11 \times 10^{- 31}

kg and

1.67 \times 10^{- 27}

kg).

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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 =1.6725×10−27kg, Mass of proton =1.6725×10−27kg ,Mass of electron =9×10−31kg)

The correct option is A 0.50625MeVn→p+eΔm=(mp+me)−mnΔm=−9×10−31kgEnergy released =9×10−31×(3×108)2J =9×10−31×(3×108)21.6×10−13 =0.50625 MeV

Assume that a neutron breaks into a proton and an electron. The energy released during this process is
(Mass of neutron $= 1.6725 \times 10^{- 27} k g$ , Mass of proton $= 1.6725 \times 10^{- 27} k g$ ,Mass of electron $= 9 \times 10^{- 31} k g$ )

The correct option is A $0.50625$ $M e V$
$n \to p + e$
$Δ m = (m_{p} + m_{e}) - m_{n}$
$Δ m = - 9 \times 10^{- 31} k g$
Energy released $= 9 \times 10^{- 31} \times (3 \times 10^{8})^{2} J$
$= \frac{9 \times 10^{- 31} \times (3 \times 10^{8})^{2}}{1.6 \times 10^{- 13}}$
$= 0.50625 M e V$