Question

The longest wavelength (in cm) of light that is required to remove an electron from n = 2 orbit of hydrogen atom is $y\times {10}^{-8}$ m. What is the value of y?

Answer 1

Here,
$\Delta E=E_f-E_i$
$=E_{\infty }-E_2$
We know that,
$E_n=-13.6\frac{Z^2}{n^2}$ eV
$\therefore \Delta E=0-(-13.6\frac{1^2}{2^2})eV=\frac{13.6}{4}eV=3.4eV$
Thus, $\Delta E=3.4eV$

Energy of the photon absorbed during this transition = $\Delta E=\frac{12400}{\lambda }$
where $\lambda$ is the wavelength and Energy E is in eV
$\Rightarrow \lambda =\frac{hc}{\Delta E}$
$\Rightarrow \lambda =\frac{12400}{3.4}$
$=3647\stackrel{o}{\text{A}}$
Now, 3647 $\stackrel{o}{\text{A}}$= 36.47$\times {10}^{-8}m$