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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Physics
Chapter
Nuclear Physics

JEE Advanced 2019 · Paper 2 · Physics · Nuclear Physics

Suppose a Ra 88226 nucleus at rest and in ground state undergoes α...

Suppose a Ra88226Ra 88226 nucleus at rest and in ground state undergoes α\alpha -decay to a Rn86222Rn 86222 nucleus in its excited state. The kinetic energy of the emitted α\alpha particle is found to be 4.44MeV.Rn862224.44MeV . Rn 86222 nucleus then goes to its ground state by γ\gamma -decay. The energy of the emitted γ\gamma -photon is _______ keV,keV,
[Given: atomic mass of 88226Ra=226.005u,{}_{88}^{226}Ra = 226.005u, atomic mass of 86222Rn=222.000u,{}_{86}^{222}Rn = 222.000u, atomic mass of α\alpha particle =4.000u,1u=931MeV/c2,c= 4.000u, 1u = 931MeV / {c}^{2},c is speed of the light]

Correct answer

0

Step-by-step solution

⇒⇒ Mass defect ∆m=226.005−222.000−4.000(Ra88226→α−decayRn+24He+γ86222)∆ m = 226.005 - 222.000 - 4.000\left(Ra 88226 \overset{\alpha - decay}{\to} Rn + {}_{2}^{4}{H}_{e} + \gamma 86222\right)
=0.005amu= 0.005amu
∴Qvalue=0.005×931.5=4.655MeV∴ Qvalue = 0.005 \times 931.5 = 4.655MeV
Also K.EαK.ERn=mRnmα\dfrac{K . {E}_{\alpha}}{K . {E}_{Rn}} = \dfrac{{m}_{Rn}}{{m}_{\alpha}}
⇒K.ERn=mαmRn.K.Eα=4222×4.44=0.08MeV⇒ K . {E}_{Rn} = \dfrac{{m}_{\alpha}}{{m}_{Rn}} . K . {E}_{\alpha} = \dfrac{4}{222} \times 4.44 = 0.08MeV
∴∴ Energy of γ−Photon=4.655−(4.44+0.08)(Eγ=Q−(KEα+KERn))\gamma - Photon = 4.655 - \left(4.44 + 0.08\right)\left({E}_{\gamma} = Q - \left(K{E}_{\alpha} + K{E}_{Rn}\right)\right)
=0.135MeV=135KeV= 0.135MeV = 135KeV
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