Modern Physics Ques 56
- 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.)
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Answer:
Correct Answer: 56.$(2)$
Solution:
Formula:
- 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 $
BETA
