Question

A proton captures a free electron whose K.E. is zero & forms a hydrogen atom of lowest energy-level (n = 1). If a photon is emitted in this process, what will be the wavelength (in $\stackrel{\circ }{A}$) of radiation? (report the answer to the nearest integer) (Ionisation potential of hydrogen =13.6 volt, $e=1.6\times {10}^{-19}J,h=6.6\times {10}^{-34}Js,c=3.0\times {10}^{8}m/s$)

Answer 1

Ionisation potential = $13.6V$
therefore, ioniation energy = $13.6eV$

Ionisation energy is given as :
$IE=E_{\infty }-E_1=\Delta E$
$13.6=0-E_1$
$E_1=-13.6eV$ (energy in ground state i.e. n=1)
$△E=\frac{hc}{\lambda }=13.6eV=13.6\times 1.6=21.76\times {10}^{-19}J$

$21.76\times {10}^{-19}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{\lambda }$
$\Rightarrow \lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{21.76\times {10}^{-19}}=909.92\stackrel{\circ }{A})\approx 910\stackrel{\circ }{A})$