JEE MAIN - Physics (2012 - No. 3)
Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron $$ = 1.6725 \times {10^{ - 27}}kg,$$ mass of proton $$ = 1.6725 \times {10^{ - 27}}\,kg,$$ mass of electron $$ = 9 \times {10^{ - 31}}\,kg$$ ).
$$0.51$$ $$MeV$$
$$7.10\,MeV$$
$$6.30\,MeV$$
$$5.4\,MeV$$
Explanation
$${}_0^1n \to {}_1^1H + {}_{ - 1}{e^0} + \overrightarrow v + Q$$
The mass defect during the process
$$\Delta m = {m_n} - {m_H} - {m_e}$$
$$ = 1.6725 \times {10^{ - 27}} - \left( {1.6725 \times {{10}^{ - 27}} + 9 \times {{10}^{ - 31}}kg} \right)$$
$$ = - 9 \times {10^{ - 31}}kg$$
The energy released during the process
$$E = \Delta m{c^2}$$
$$E = 9 \times {10^{ - 31}} \times 9 \times {10^{16}}$$
$$ = 81 \times {10^{ - 15}}\,joules$$
$$E = {{81 \times {{10}^{ - 15}}} \over {1.6 \times {{10}^{ - 19}}}} = 0.511MeV$$
The mass defect during the process
$$\Delta m = {m_n} - {m_H} - {m_e}$$
$$ = 1.6725 \times {10^{ - 27}} - \left( {1.6725 \times {{10}^{ - 27}} + 9 \times {{10}^{ - 31}}kg} \right)$$
$$ = - 9 \times {10^{ - 31}}kg$$
The energy released during the process
$$E = \Delta m{c^2}$$
$$E = 9 \times {10^{ - 31}} \times 9 \times {10^{16}}$$
$$ = 81 \times {10^{ - 15}}\,joules$$
$$E = {{81 \times {{10}^{ - 15}}} \over {1.6 \times {{10}^{ - 19}}}} = 0.511MeV$$
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