Question

According to Bohr theory, the electronic energy of the hydrogen atom in the $n^{th}$ Bohr orbit is given by $E_n=-\frac{21\times {10}^{-19}}{n^2}\times Z^{2}J$
The longest wavelength of light that will be needed to remove an electron from the $3^{rd}$ orbit of the $H{e}^{+}$ ion is $x\times {10}^{-7}m$. Find the value of $x$.
(Give your answer upto two decimal only, take $h=6\times {10}^{-34}Jsec$)

Answer 1

The electronic energy of $H{e}^{+}$ ion in the $n^{th}$ Bohr orbit =$\frac{-21\times {10}^{-19}}{n^2}\times Z^{2}J$
Where,$Z=2$
Thus, energy of $H{e}^{+}$ in the $3^{rd}$ Bohr orbit
$=\frac{-21\times {10}^{-19}}{3^2}\times 2^{2}J$
$=\frac{-21\times {10}^{-19}}{9}\times 4J$
$\Delta E=E_{\infty }-E_3$
$=0-[\frac{-21\times {10}^{-19}}{9}\times 4]$
$=\frac{21\times {10}^{-19}}{9}\times 4$
We know that,
$\lambda =\frac{hc}{\Delta E}=\frac{6\times {10}^{-34}\times 3\times {10}^8\times 9}{21\times {10}^{-19}\times 4}$
$=1.92\times {10}^{-7}m$