Modern Physics 1 Question 48
44. Consider a hydrogen atom with its electro in the $n^{\text {th }}$ orbital. An electromagnetic radiation of wavelength $90 nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 eV$, then the value of $n$ is $(h c=1242 eV nm)$ (2015 Adv.)
Analytical & Descriptive Questions
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Solution:
- Kinetic energy of ejected electron
$=$ Energy of incident photon - energy required to ionize the electron from $n$th orbit (all in $eV$ )
$$ \begin{aligned} \therefore \quad 10.4 & =\frac{1242}{90}-\left|E _n\right| \\ & =\frac{1242}{90}-\frac{13.6}{n^{2}} \quad\left(\text { as } E _n \propto \frac{1}{n^{2}} \text { and } E _1=-13.6 eV\right) \end{aligned} $$
Solving this equation, we get
$$ n=2 $$
