Question

According to Bohr theory, the electronic energy of the hydrogen atom in the $n$th Bohr orbit is given by
$E_n=-\frac{21.76\times {10}^{-19}}{n^2}J$
Calculate the longest wavelength of light that will be needed to remove an electron from the $3$rd orbit of the ${He}^{+}$ ion.

Answer 1

The electronic energy of ${He}^{+}$ ion in the $n$th Bohr orbit
$=-\frac{21.76\times {10}^{-19}}{n^2}\times Z^{2}J$
where, $Z=2$
Thus, energy of ${He}^{+}$ in the $3$rd Bohr orbit
$=-\frac{21.76\times {10}^{-19}}{9}\times 4J$
$\Delta E=E_{\infty }-E_3$
$=0-[-\frac{21.76\times {10}^{-19}\times 4}{9}]$
$=\frac{21.76\times {10}^{-19}\times 4}{9}$
We know that, $\lambda =\frac{hc}{\Delta E}=\frac{6.625\times {10}^{-34}\times 3\times {10}^8\times 9}{21.76\times {10}^{-19}\times 4}$
$=2.055\times {10}^{-7}m$.