Question

Estimate the difference in energy between $1st$ and $2nd$ Bohr’s orbit for a hydrogen atom. At what minimum atomic number, a transfer from $n=2$ to $n=1$ energy level would result in the emission of X-rays with $\lambda =3.0\times {10}^{–8}m$ ? Which among the following hydrogen atom-like species does this atomic number correspond to:
(use value of $R_H=1.097\times {10}^7{m}^{-1})$

• A. $Z=3i.e.L{i}^{2+}ion$
• B. $Z=2i.e.H{e}^{+}ion$
• C. $Z=4i.e.B{e}^{3+}ion$
• D. None of the above

Answer 1

The correct option is B $Z=2i.e.H{e}^{+}ion$

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