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  • The ratio of nucleus recoil energy in <img alt="\mathrm{ \alpha \text {-decay of }{ }^{226} \mathrm{Ra} \text { to that of in its } \gamma \text { - }}" src="https://entrancecorner.oncodecogs.

The ratio of nucleus recoil energy in \mathrm{ \alpha \text {-decay of }{ }^{226} \mathrm{Ra} \text { to that of in its } \gamma \text { - }} decay if the total energy released in α -decay is \mathrm{E_1} = 4.9 MeV and that released in the\mathrm{\gamma \text {-decay is } E_2=0.2 \mathrm{MeV} \text { is found } y \times 10^5 \text {. }} Then value of y is:

Option: 1

4


Option: 2

6


Option: 3

7


Option: 4

9


Answers (1)

best_answer

According to the law of conservation of momentum in the case of \gamma-decay we have

\mathrm{\mathrm{Mv}_2=\frac{E_v}{c} .}

\mathrm{\text { Here } M=226 \times 1.66 \times 10^{-27} \mathrm{~kg}}

\mathrm{=3.75 \times 10^{-25} \mathrm{~kg}} is the mass of \mathrm{\text { a }{ }^{226} \text { Ra nucleus, } E_\alpha \text { the } \gamma \text {-quantum }}energy practically coincident with the total energy \mathrm{E_2 \text { released in the } \gamma \text {-decay, }}, and c the velocity of light in vacuum. Whence for the kinetic energy \mathrm{T_2} of the
recoil nucleus we have,

\mathrm{T_2=\frac{E_2^2}{2 M c^2}=0.095 \mathrm{eV}}

In the case of \mathrm{\alpha \text {-decay }} the laws of conservation or energy and momentum yield

\mathrm{\begin{aligned} & \left(M-M_\alpha\right) v_1=M_\alpha v_\alpha \\ & \quad \frac{1}{2}\left(M-M_\alpha\right) v_1^2+\frac{1}{2} M_\alpha v_\alpha^2=E_1 \end{aligned}}

where \mathrm{\boldsymbol{V}_\alpha \text { and } \boldsymbol{v}_1} are the velocities of the \mathrm{\alpha}-particle and recoil nucleus, respectively, and \mathrm{M_\alpha} the \mathrm{\alpha}-particle mass. From where, by eliminating \mathrm{V_\alpha}, we
obtain the following

\mathrm{T_1=\frac{M_{\mathrm{a}}}{M} E_1=87 \mathrm{keV} .}

Finally, the ratio of the kinetic energies is

\mathrm{\frac{T_1}{T_2}=\frac{2 M_{\mathrm{a}} C^2 E_1}{E_2^2}=9 \times 10^5 .}

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Gunjita

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