JEE Physics 2023 Question with Solution

Given: The transition energies are $\Delta E_A = 2.2 \, \text{eV}$, $\Delta E_B = 5.2 \, \text{eV}$, $\Delta E_C = 3 \, \text{eV}$, and $\Delta E_D = 10 \, \text{eV}$.

Find: Which transition emits a photon of wavelength $124.1 \, \text{nm}$.

Use

$$\lambda = \frac{hc}{\Delta E}$$

For the listed transitions:

$$\lambda_A = \frac{1241}{2.2} \, \text{nm} = 564 \, \text{nm}$$ $$\lambda_B = \frac{1241}{5.2} \, \text{nm} = 238.65 \, \text{nm}$$ $$\lambda_C = \frac{1241}{3} \, \text{nm} = 413.66 \, \text{nm}$$ $$\lambda_D = \frac{1241}{10} \, \text{nm} = 124.1 \, \text{nm}$$

This matches the given wavelength.

Therefore, the correct transition is D.

Given: $h = 6.626 \times 10^{-34} \, \text{J s}$, $c = 3 \times 10^8 \, \text{m/s}$ and for transition D, $\Delta E_D = 10 \, \text{eV}$.

Find: Whether this gives $124.1 \, \text{nm}$.

Using

$$\lambda = \frac{hc}{\Delta E}$$

Substitute $\Delta E = 10 \times 1.6 \times 10^{-19} \, \text{J}$:

$$\lambda_D = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10 \times 1.6 \times 10^{-19}} \, \text{m}$$

This gives

$$\lambda_D = 124.1 \, \text{nm}$$

Hence the emitted photon corresponds to transition D.

The solution also contains a discrepancy where one heading says the correct option is A, but the worked values clearly show the correct transition is D. Since the option list is arranged as A: B, B: A, C: C, D: D, the correct option is D.