JEE PYQ: Atomic Physics Question 47
Question 47 - 2019 (08 Apr Shift 1)
Radiation coming from transitions $n = 2$ to $n = 1$ of hydrogen atoms fall on He$^+$ ions in $n = 1$ and $n = 2$ states. The possible transition of helium ions as they absorb energy from the radiation is:
(1) $n = 2 \to n = 3$
(2) $n = 1 \to n = 4$
(3) $n = 2 \to n = 5$
(4) $n = 2 \to n = 4$
Show Answer
Answer: (4)
Solution
Energy released by hydrogen atom for transition from $n = 2$ to $n = 1$: $\Delta E_1 = 13.6 \times (\frac{1}{1^2} - \frac{1}{2^2}) = \frac{3}{4} \times 13.6 = 10.2$ eV. This energy is absorbed by He$^+$ ion in transition from $n = 2$ to $n = n_1$ (say). $\Delta E_2 = 13.6 \times 4 \times \left(\frac{1}{4} - \frac{1}{n_1^2}\right) = 10.2$ eV. So $n_1 = 4$. Possible transition is $n = 2 \to n = 4$.
BETA
