Question

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

Answer 1

For H-atom, the energy of a stationary orbit is determined as
$E_n=-\frac{k}{n^2}$ where, $k=$ constant $\left(2.18\times 10^{-18}J\right)$
$\Rightarrow \Delta E(n=2\text{ to }n=1)=k\left(1-\frac{1}{4}\right)=\frac{3}{4}k$
$=1.635\times 10^{-18}J$
For a H-like species, energy of stationary orbit is determined as
$E_n=-\frac{k{Z}^2}{n^2}$
where, $Z=$ atomic number
$\Rightarrow \Delta E=k{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
$\Rightarrow \frac{1}{\lambda }=\frac{\Delta E}{hc}=\frac{k}{hc}{Z}^2\left(\frac{1}{1}-\frac{1}{4}\right)=R_H{Z}^2\times \frac{3}{4}$
$\Rightarrow Z^2=\frac{4}{3R_H\lambda }=\frac{4}{3\times 1.097\times 10^7\times 3\times 10^{-8}}=4.05$
$\Rightarrow Z=2\left(H{e}^{+}\right)$