Modern Physics Ques 50
- Consider a hydrogen-like ionised atom with atomic number $Z$ with a single electron. In the emission spectrum of this atom, the photon emitted in the $n=2$ to $n=1$ transition has energy $74.8 $ $eV$ higher than the photon emitted in the $n=3$ to $n=2$ transition. The ionisation energy of the hydrogen atom is $13.6$ $eV$. The value of $Z$ is
(2018 Adv.)
Show Answer
Answer:
Correct Answer: 50.$(3)$
Solution:
Formula:
- $\Delta E _{2-1}=13.6 \times Z^{2} \left[1-\frac{1}{4}\right]=13.6 \times Z^{2} \left[\frac{3}{4}\right]$
$ \begin{aligned} & \Delta E _{3-2}=13.6 \times Z^{2} \left[\frac{1}{4}-\frac{1}{9}\right]=13.6 \times Z^{2} \left[\frac{5}{36}\right] \\ & \therefore \quad \Delta E _2=\Delta E _{3-2}+74.8 \\ & 13.6 \times Z^{2} \left[\frac{3}{4}\right]=13.6 \times Z^{2} \left[\frac{5}{36}\right]+74.8 \\ & 13.6 \times Z^{2} \left[\frac{3}{4}-\frac{5}{36}\right]=74.8 \\ & Z^{2}=9 \\ & Z=3 \end{aligned} $
BETA
