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AIEEE 2012 Solved Paper

Section: Physics

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Question : 55 of 89

Marks: +1, -0

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} \text{kg}

, mass of proton

=1.6725 \times 10^{-27} \text{kg}

, mass of electron

=9 \times 10^{-31} \text{kg})

$0.51 \text{MeV}$
$7.10 \text{MeV}$
$6.30 \text{MeV}$
$5.4 \text{MeV}$

Solution:

{}_{0}^{1}n \rightarrow {}_{1}^{1}H+{}_{-1}^{0}e+\overset{\rightarrow}{v}+Q

The mass defect during the process

\;\Delta m=m_n-m_H-m_e

\;=1.6725 \times 10^{-27} - (1.6725 \times 10^{-27} + 9 \times 10^{-31} \text{kg})

\;=-9 \times 10^{-31} \text{kg}

The energy released during the process

\;E=\Delta m c^2

\;E=9 \times 10^{-31} \times 9 \times 10^{16}

\;=81 \times 10^{-15} \;\text{joules}\;

\;E=\;\frac{81 \times 10^{-15}}{1.6 \times 10^{-19}}=0.511 \text{MeV}

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