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PhysicsNumerical+3Question 13 of 54

JEE Advanced 2019 · Paper 2, Physics question 13

Suppose a X88226X2882226Ra\ce{^{226}_{88}Ra} nucleus at rest and in ground state undergoes α-decay to a X86222X2862222Rn\ce{^{222}_{86}Rn} nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 MeV. X86222X2862222Rn\ce{^{222}_{86}Rn} nucleus then goes to its ground state by γ-decay. The energy of the emitted γ photon is ______ keV.

[Given: atomic mass of X88226X2882226Ra\ce{^{226}_{88}Ra} = 226.005 u, atomic mass of X86222X2862222Rn\ce{^{222}_{86}Rn} = 222.000 u, atomic mass of α particle = 4.000 u, 1 u = 931 MeV/c², c is speed of the light]

Answer and worked solution

Official answer: 130

Energy released.
Q=(226.005−222.000−4.000)×931=4.655 MeV.Q=(226.005-222.000-4.000)\times931=4.655\text{ MeV}.

Recoil. Momentum is conserved, so the Rn nucleus recoils with
KRn=KαmαmRn=4.44×4222=0.080 MeV.K_{Rn}=K_\alpha\frac{m_\alpha}{m_{Rn}}=4.44\times\frac4{222}=0.080\text{ MeV}.

Gamma energy.
Eγ=4.655−4.44−0.080=0.135 MeV=135 keV.E_\gamma=4.655-4.44-0.080=0.135\text{ MeV}=135\text{ keV}.
Answer: 135 (key: 130 to 140)

Topics:Mass defect and binding energyConservation of linear momentumRadioactive decay
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