Question

According to Bohr's theory, the energy of an electron in the $n^{th}$ energy level of hydrogen like species is given by:
$E_n=\frac{-21.76\times {10}^{-19}}{n^2}\left(Z^2\right)J$
Calculate the longest wavelength of light that will be needed to completely remove an electron from the third Bohr orbit of $H{e}^{+}$ ion.

• A. $2055\stackrel{\circ }{\text{A}}$
• B. $205.5\stackrel{\circ }{\text{A}}$
• C. $20.55\stackrel{\circ }{\text{A}}$
• D. $2.055\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is A $2055\stackrel{\circ }{\text{A}}$

We know that, for any $H$ like species, $E_n=\frac{-21.76\times {10}^{-19}}{n^2}\left(Z^2\right)$
where, $n=\text{Orbit number},Z=\text{Atomic number}$
For $H{e}^{+}$ ion $(Z=2)$ in $3^{rd}$ orbit,
$E_3\left(H{e}^{+}\right)=\frac{-21.76\times {10}^{-19}}{3^2}\left(2^2\right)\text{J}$
$=\frac{-21.76\times {10}^{-19}}{9}\left(4\right)\text{J}...\left(i\right)$

Hence, the energy equivalent to $E_3$ (magnitude wise) will be required to remove an $e^{-}$ from $3^{rd}$ orbit of $H{e}^{+}$ ion.
Also, $E=\frac{hc}{\lambda }...\left(ii\right)$

Hence, equating (i) and (ii) we get,
$\frac{hc}{\lambda }=\frac{21.76\times {10}^{-19}}{9}\left(4\right)$
$\Rightarrow \lambda =\frac{hc}{E}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8\times 9}{21.76\times {10}^{-19}\times 4}$

$\Rightarrow \lambda =2055\times {10}^{-10}\text{m}$
$\Rightarrow \lambda =2055\stackrel{\circ }{\text{A}}$