Question

Energy of an electron is given by $E=-2.178\times {10}^{18}J\left(\frac{Z^2}{n^2}\right).$
Wavelength of light required to excite an electron in an hydrogen atom from level $n=1$ to $n=2$ will be:
$(h=6.62\times {10}^{-34}Js$ and $c=3.0\times {10}^{8}m{s}^{-1})$

• A. $6.500\times {10}^{-7}m$
• B. $8.500\times {10}^{-7}m$
• C. $1.214\times {10}^{-7}m$
• D. $2.816\times {10}^{-7}m$

Answer 1

Answer: (C)

$1.214\times {10}^{-7}m$

Let $\lambda$ is the wavelength of that photon then
$hc/\lambda =E_2-E_1$
and using value of given E,
$E=\frac{hc}{\lambda }=2.178\times {10}^{-18}\times z^2[\frac{1}{1^2}-\frac{1}{2^2}](z=1)$
$\therefore \lambda =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{2.178\times {10}^{-18}}\times \frac{4}{3}=1.214\times {10}^{-7}m$