Search any question & find its solution
Question:
Answered & Verified by Expert
Obtain the maximum kinetic energy of $\beta$-particles, and the radiation frequencies of $\gamma$ decys in the decay scheme shown in fig. You are given that
$$
\mathrm{m}\left({ }^{198} \mathrm{Au}\right)=197.968233 \mathrm{u} ; \mathrm{m}\left({ }^{198} \mathrm{Hg}\right)=197.966760 \mathrm{u}
$$
$$
\mathrm{m}\left({ }^{198} \mathrm{Au}\right)=197.968233 \mathrm{u} ; \mathrm{m}\left({ }^{198} \mathrm{Hg}\right)=197.966760 \mathrm{u}
$$
Solution:
2460 Upvotes
Verified Answer
Energy corresponding to $\gamma_1$
$\mathrm{E}_1=1.088-0=1.088 \mathrm{MeV}=1.088 \times 1.6 \times 10^{-13} \mathrm{~J}$
$\therefore$ Frequency $\nu_1$ for $\gamma_1=\frac{E_1}{h}$
$=\frac{1.088 \times 1.6 \times 10^{-13}}{6.6 \times 10^{-34}} \cong 2.63 \times 10^{20} \mathrm{~Hz}$
Similarly $\nu_2$, for $\gamma_2$
$=\frac{E_2}{h}=\frac{0.412 \times 1.6 \times 10^{-13}}{6.6 \times 10^{-34}}=9.98 \times 10^{19} \mathrm{~Hz}$ and
$\nu_3$ for $\gamma_3=\frac{E_3}{h}=\frac{(1.088-0.412) \mathrm{MeV}}{\mathrm{h}}$
$=\frac{(1.088-0.412) \times 1.6 \times 10^{-13}}{6.6 \times 10^{-34}}$
$=1.64 \times 10^{20} \mathrm{~Hz}$
Maximum $\mathrm{K}$. E of $\beta_1$ particle, $\mathrm{K}_{\max } \beta_1$
$=\left[\mathrm{m}\left({ }_{79}^{198} \mathrm{Au}\right)-\mathrm{m}\left({ }_{80}^{198} \mathrm{Hg}\right)-\frac{1.088}{931}\right]$
$=(197.968233-197.966760) \times 931-1.088$
$-1.088=1.371-1.088=0.283 \mathrm{MeV}$
Similarly, $\mathrm{K}_{\text {max }}\left(\beta_2\right)=$
$\left[\mathrm{m}\left({ }_{79}^{198} \mathrm{Au}\right)-\mathrm{m}\left({ }_{80}^{198} \mathrm{Hg}\right)-\mathrm{m}\left(\frac{0.412}{931}\right)\right] \times 931$
$=(197.968233-197.966760) \times 931-0.412$
$=0.957 \mathrm{MeV}$.
$\mathrm{E}_1=1.088-0=1.088 \mathrm{MeV}=1.088 \times 1.6 \times 10^{-13} \mathrm{~J}$
$\therefore$ Frequency $\nu_1$ for $\gamma_1=\frac{E_1}{h}$
$=\frac{1.088 \times 1.6 \times 10^{-13}}{6.6 \times 10^{-34}} \cong 2.63 \times 10^{20} \mathrm{~Hz}$
Similarly $\nu_2$, for $\gamma_2$
$=\frac{E_2}{h}=\frac{0.412 \times 1.6 \times 10^{-13}}{6.6 \times 10^{-34}}=9.98 \times 10^{19} \mathrm{~Hz}$ and
$\nu_3$ for $\gamma_3=\frac{E_3}{h}=\frac{(1.088-0.412) \mathrm{MeV}}{\mathrm{h}}$
$=\frac{(1.088-0.412) \times 1.6 \times 10^{-13}}{6.6 \times 10^{-34}}$
$=1.64 \times 10^{20} \mathrm{~Hz}$
Maximum $\mathrm{K}$. E of $\beta_1$ particle, $\mathrm{K}_{\max } \beta_1$
$=\left[\mathrm{m}\left({ }_{79}^{198} \mathrm{Au}\right)-\mathrm{m}\left({ }_{80}^{198} \mathrm{Hg}\right)-\frac{1.088}{931}\right]$
$=(197.968233-197.966760) \times 931-1.088$
$-1.088=1.371-1.088=0.283 \mathrm{MeV}$
Similarly, $\mathrm{K}_{\text {max }}\left(\beta_2\right)=$
$\left[\mathrm{m}\left({ }_{79}^{198} \mathrm{Au}\right)-\mathrm{m}\left({ }_{80}^{198} \mathrm{Hg}\right)-\mathrm{m}\left(\frac{0.412}{931}\right)\right] \times 931$
$=(197.968233-197.966760) \times 931-0.412$
$=0.957 \mathrm{MeV}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.