JEE PYQ: Atomic Physics Question 55
Question 55 - 2019 (12 Apr Shift 1)
An excited He$^+$ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number $n$, corresponding to its initial excited state is (for photon of wavelength $\lambda$, energy $E = \frac{1240 \text{eV}}{\lambda (\text{nm})}$):
(1) $n = 4$
(2) $n = 5$
(3) $n = 7$
(4) $n = 6$
Show Answer
Answer: (2)
Solution
$E = E_1 + E_2 = \frac{1240}{\lambda_1} + \frac{1240}{\lambda_2}$. $\frac{13.6(2)^2}{n^2} = 1240\left(\frac{1}{108.5} + \frac{1}{30.4}\right) \times \frac{1}{10^{-9}}$. On solving, $n = 5$.
BETA
