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
$\Rightarrow \Delta E=k Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
$\Rightarrow \frac{1}{\lambda}=\frac{\Delta E}{h c}=\frac{k}{h c} Z^{2}\left(\frac{1}{1}-\frac{1}{4}\right)=R_{ H } Z^{2} \times \frac{3}{4}$
$\Rightarrow Z^{2}=\frac{4}{3 R_{ H } \lambda}=\frac{4}{3 \times 1.097 \times 10^{7} \times 3 \times 10^{-8}}=4.05$
$\Rightarrow Z=2\left( He ^{+}\right)$