Question

Let the energy of an $ n^{th} $ orbit of $ H $ atom be $ -21.76 \times 10^{19}/n^2\, J $ . What will be the longest wavelength of energy required to remove an electron from the third orbit?

• A. $ 0.628\, nm $
• B. $ 1.326 \times 10^{-7}\, m $
• C. $ 0.798\, pm $
• D. $ 0.821 \,\mu m $

Answer 1

Answer: (D)

$E_{n}=\frac{-21.76\times10^{-19}}{n^{2}} J $
For $n = 3$
$E_{n}=\frac{-21.76\times10^{-19}}{\left(3\right)^{2}}$
$=-2.42\times10^{-19} J$
$E=\frac{hc}{\lambda}$
$\Rightarrow \lambda=\frac{hc}{E}$
$=\frac{6.6\times10^{-34}\times3\times10^{8}}{2.42\times10^{-19}}$
$=8.2\times10^{-7}\,m$
$=0.82\times10^{-6}\,m$
$=0.82\, \mu m$