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JEE Main 2012Offline
Exam
JEE Main
Year
2012
Mode
Offline
Subject
Physics
Chapter
Nuclear Physics

JEE Main 2012 · Offline · Physics · Nuclear Physics

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

Options

  1. A0.73 MeV0.73 \mathrm{~MeV}
  2. B7.10 MeV7.10 \mathrm{~MeV}
  3. C6.30 MeV6.30 \mathrm{~MeV}
  4. D5.4 MeV5.4 \mathrm{~MeV}

Correct answer

A. 0.73 MeV0.73 \mathrm{~MeV}

Step-by-step solution

Δm=(mp+me)−mn=9×10−31 kg.\begin{aligned} \Delta m & =\left(m_p+m_e\right)-m_n \\ & =9 \times 10^{-31} \mathrm{~kg}. \end{aligned} Energy released =(9×10−31 kg)c2=\left(9 \times 10^{-31} \mathrm{~kg}\right) \mathrm{c}^2 joules =9×10−31×(3×108)21.6×10−13MeV=0.73 MeV.\begin{aligned} & =\frac{9 \times 10^{-31} \times\left(3 \times 10^8\right)^2}{1.6 \times 10^{-13}} \mathrm{MeV} \\ & =0.73 \mathrm{~MeV} . \end{aligned}
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