Verified Solution • 2025–26 Session moderate

QJEEM12Ph1323: JEE Main 2012 Physics - Class 12 JEEM12Ph1323 Solution: Assume that a neutron breaks into a proton and an electron. The …

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,= 1.6725 \times {10^{ - 27}}kg, mass of proton =1.6725×10−27 kg,= 1.6725 \times {10^{ - 27}}\,kg, mass of electron =9×10−31 kg= 9 \times {10^{ - 31}}\,kg ).

Multiple Choice Options

  • A 0.510.51 MeVMeV
  • B 7.10 MeV7.10\,MeV
  • C 6.30 MeV6.30\,MeV
  • D 5.4 MeV5.4\,MeV

Step-by-Step Verified Solution

01n→11H+−1e0+v→+Q{}_0^1n \to {}_1^1H + {}_{ - 1}{e^0} + \overrightarrow v + Q

The mass defect during the process

Δm=mn−mH−me\Delta m = {m_n} - {m_H} - {m_e}

=1.6725×10−27−(1.6725×10−27+9×10−31kg)= 1.6725 \times {10^{ - 27}} - \left( {1.6725 \times {{10}^{ - 27}} + 9 \times {{10}^{ - 31}}kg} \right)

=−9×10−31kg= - 9 \times {10^{ - 31}}kg

The energy released during the process

E=Δmc2E = \Delta m{c^2}

E=9×10−31×9×1016E = 9 \times {10^{ - 31}} \times 9 \times {10^{16}}

=81×10−15 joules= 81 \times {10^{ - 15}}\,joules

E=81×10−151.6×10−19=0.511MeVE = {{81 \times {{10}^{ - 15}}} \over {1.6 \times {{10}^{ - 19}}}} = 0.511MeV

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