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An energy of 68.0 eV is required to excite a hydrogen-like atom in its second Bohr energy level to third energy level the charge of nucleus is Ze . The wavelength of a radiation required to eject the electron from first orbit to infinity is
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2.5 nm
Given, $n_1=2, n_2=3$ and $\Delta E=68 \mathrm{eV}$
The difference in energies of two orbits
$\Delta E=13.6 Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
where, $n_1 \lt n_2$
$\begin{aligned} 68 & =13.6 Z^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ & =13.6 Z^2\left[\frac{1}{4}-\frac{1}{9}\right] \\ & =13.6 Z^2\left[\frac{9-4}{36}\right] \\ & =13.6 Z^2 \times \frac{5}{36}\end{aligned}$
$\begin{aligned} \therefore \quad Z^2 & =\frac{68 \times 36}{13.6 \times 5}=36 \\ Z & =6\end{aligned}$
$\because$ Wavelength of photon, $\frac{1}{\lambda}=Z^2 R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$n_1=1$ and $n_2=\infty$
$\frac{1}{\lambda}=6^2 \times R\left[\frac{1}{1^2}-\frac{1}{\infty}\right]=36 R$
$\therefore \quad \lambda=\frac{1}{36 R}=\frac{1}{36 \times 1.097 \times 10^7}$
$\begin{aligned} & =\frac{10^{-7}}{39.5} \mathrm{~m} \\ & =0.025 \times 10^{-7} \mathrm{~m} \\ & =2.5 \mathrm{~nm}\end{aligned}$
The difference in energies of two orbits
$\Delta E=13.6 Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
where, $n_1 \lt n_2$
$\begin{aligned} 68 & =13.6 Z^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ & =13.6 Z^2\left[\frac{1}{4}-\frac{1}{9}\right] \\ & =13.6 Z^2\left[\frac{9-4}{36}\right] \\ & =13.6 Z^2 \times \frac{5}{36}\end{aligned}$
$\begin{aligned} \therefore \quad Z^2 & =\frac{68 \times 36}{13.6 \times 5}=36 \\ Z & =6\end{aligned}$
$\because$ Wavelength of photon, $\frac{1}{\lambda}=Z^2 R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$n_1=1$ and $n_2=\infty$
$\frac{1}{\lambda}=6^2 \times R\left[\frac{1}{1^2}-\frac{1}{\infty}\right]=36 R$
$\therefore \quad \lambda=\frac{1}{36 R}=\frac{1}{36 \times 1.097 \times 10^7}$
$\begin{aligned} & =\frac{10^{-7}}{39.5} \mathrm{~m} \\ & =0.025 \times 10^{-7} \mathrm{~m} \\ & =2.5 \mathrm{~nm}\end{aligned}$
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