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QJEEAdv19Ph2cbc: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph2cbc Solution: Suppose a nucleus at rest and in ground state undergoes -decay…

Suppose a 88226Ra_{88}^{226}Ra nucleus at rest and in ground state undergoes α\alpha -decay to a 86222Rn_{86}^{222}Rn nucleus in its excited state. The kinetic energy of the emitted α\alpha particle is found to be 4.44 MeV. 86222Rn_{86}^{222}Rn nucleus then goes to its ground state by γ\gamma -decay. The energy of the emitted γ\gamma photon is ............ keV.

[Given : atomic mass of 86226Ra_{86}^{226}Ra = 226.005 u, atomic of 86222Rn_{86}^{222}Rn = 222.000 u, atomic mass of α\alpha particle = 4.000 u, 1 u = 931 MeV/e2, c is speed of the light]

Step-by-Step Verified Solution

Given, atomic mass of 88226Ra=226.005u{ }_{88}^{226} \mathrm{R} a=226.005 u ,

atomic mass of 86222Rn=222.000u{ }_{86}^{222} \mathrm{R} n=222.000 u .

atomic mass of α\alpha particle =4.000u=4.000 u

1u=931MeVc21 u=931 \frac{\mathrm{MeV}}{c^2}

where, c = speed of the light

88226Ra⟶α− decay 86222Rn+24He{ }_{88}^{226} \mathrm{R} a \stackrel{\alpha-\text { decay }}{\longrightarrow}{ }_{86}^{222} \mathrm{R} n+{ }_2^4 \mathrm{He}

Total energy emitted =(Δm)c2=(\Delta m) c^2

amp;=[226.005−(222+14)]×931.5Mevamp;=0.005×931.5MeV\begin{aligned}& =[226.005-(222+14)] \times 931.5 \mathrm{Mev} \\\\& =0.005 \times 931.5 \mathrm{MeV}\end{aligned}

Kinetic energy of α=4.44MeV\alpha=4.44 \mathrm{MeV}

Kinetic energy of Rn = 4.44MeV×42224.44 \mathrm{MeV} \times \frac{4}{222} = 0.08 MeV

amp; Energy of photon (γ)=(Δm)c2−(k.Eα+kERn)amp;Er=[[0.05×931.5]−(4.44+0.08)]MeVamp;Er=[4.655−4.52]MeVamp;Er=0.135MeV=0.135×103keVamp;Er=135keV\begin{aligned}& \text { Energy of photon }(\gamma)=(\Delta m) c^2-\left(k . \mathrm{E}_\alpha+k \mathrm{E}_{\mathrm{R} n}\right) \\\\& \mathrm{E}_r=[[0.05 \times 931.5]-(4.44+0.08)] \mathrm{MeV} \\\\& \mathrm{E}_r=[4.655-4.52] \mathrm{MeV} \\\\& \mathrm{E}_r=0.135 \mathrm{MeV}=0.135 \times 10^3 \mathrm{keV} \\\\& \mathrm{E}_r=135 \mathrm{keV}\end{aligned}

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