JEE Physics 2025 Question with Solution

Given: The orbital radius in the ground state is $r_1 = 5.3 \times 10^{-11} \, m$ and the orbital radius in the third excited state is $r_4 = 8.48 \times 10^{-10} \, m$.

Find: The ratio $\lambda_1/\lambda_4$ of the de Broglie wavelengths.

From Bohr's quantization condition,

and the de Broglie relation,

So,

Therefore,

Since the third excited state corresponds to $n_4 = 4$ and the ground state corresponds to $n_1 = 1$,

Now,

So the physically correct ratio is $\frac{1}{4}$.

The solution contains a discrepancy because one approach marks Option A while the detailed working gives $\lambda_1/\lambda_4 = 1/4$, equivalently $\lambda_4 : \lambda_1 = 4 : 1$. Since the listed options contain only $4$, the most defensible marked option is A.

Given: $r_n = n^2 r_1$ in the Bohr model.

Find: The wavelength ratio between the ground state and the third excited state.

For the third excited state, $n = 4$. Hence,

This also matches the given radii because

Using

we get

Therefore,

Thus, the ratio of de Broglie wavelengths of the electron in the ground state to that in the excited state is $1/4$. Because the solution's labels Option A as correct despite this mismatch, the extracted answer is A with the discrepancy noted.