Verified Solution • 2025–26 Session moderate

QJEEAdv21Ph091c: JEE Advanced 2021 Physics - Class 12 JEEAdv21Ph091c Solution: A heavy nucleus N, at rest, undergoes fission N P + Q, where P…

A heavy nucleus N, at rest, undergoes fission N →\to P + Q, where P and Q are two lighter nuclei. Let δ\delta = MN −- MP −- MQ, where MP, MQ and MN are the masses of P, Q and N, respectively. EP and EQ are the kinetic energies of P and Q, respectively. The speeds of P and Q are vP and vQ, respectively. If c is the speed of light, which of the following statement(s) is(are) correct?

Multiple Choice Options

  • A EP+EQ=c2δ{E_P} + {E_Q} = {c^2}\delta
  • B EP=(MPMP+MQ)c2δ{E_P} = \left( {{{{M_P}} \over {{M_P} + {M_Q}}}} \right){c^2}\delta
  • C vPvQ=MQMP{{{v_P}} \over {{v_Q}}} = {{{M_Q}} \over {{M_P}}}
  • D The magnitude of momentum for P as well Q is c2μδc\sqrt {2\mu \delta } , where μ=MPMQ(MP+MQ)\mu = {{{M_P}{M_Q}} \over {({M_P} + {M_Q})}}

Step-by-Step Verified Solution

JEE Advanced 2021 Paper 2 Online Physics - Atoms and Nuclei Question 30 English Explanation
(ptotal) = 0 ⇒\Rightarrow (ptotal)f = 0

So, momentum of one nucleus is p in forward direction, then momentum of the other nucleus will be p in backwards direction.

Energy released = ( Δ\Delta m)c2 = δ\delta c2 = KEP + KEQ

KEP=p22mP,KEQ=p22mQK{E_P} = {{{p^2}} \over {2{m_P}}},K{E_Q} = {{{p^2}} \over {2{m_Q}}}

⇒KEP:KEQ=1mP:1mQ=mQ:mP\Rightarrow K{E_P}:K{E_Q} = {1 \over {{m_P}}}:{1 \over {{m_Q}}} = {m_Q}:{m_P}

KEP=(mQmP+mQ)δc2K{E_P} = \left( {{{{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}

KEQ=(mPmP+mQ)δc2K{E_Q} = \left( {{{{m_P}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}

KEP+KEQ=δc2K{E_P} + K{E_Q} = \delta {c^2}

p22mP+p22mQ=δc2⇒p=c2(mPmQmP+mQ)δ{{{p^2}} \over {2{m_P}}} + {{{p^2}} \over {2{m_Q}}} = \delta {c^2} \Rightarrow p = c\sqrt {2\left( {{{{m_P}{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta }

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