Question

Estimate the difference in energy between the first and second Bohr orbit for hydrogen atom. At what minimum atomic number would a transition from $n=2$ to $n=1$ energy level result in the emission of $X$-rays with $\lambda =3.0\times {10}^{-8}m$? Which hydrogen-like species does this atomic number correspond to?

Answer 1

$\Delta E=hv=\frac{h\cdot c}{\lambda }$
and $\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\Delta E=R\cdot h\cdot c[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\Delta E=h\cdot c\cdot \frac{3}{4}R$
$=\frac{6.625\times {10}^{-34}\times 3\times {10}^8\times 1.09678\times {10}^7\times 3}{4}$
$=1.635\times {10}^{-18}J$
For hydrogen-like species,
$\Delta E=Z^{2}Rhc[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{1}{\lambda }=Z^{2}R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{1}{3.0\times {10}^{-8}}=Z^2\times 1.09678\times {10}^7\times [\frac{1}{1^2}-\frac{1}{2^2}]$
$Z^2=\frac{4}{3\times {10}^{-8}\times 1.09678\times {10}^7\times 3}=4$
or $Z=2$
The species is ${He}^{+}$.