JEE Physics 2025 Question with Solution

Given: Atomic hydrogen is in the **$4$**th energy level.

Find: The number of spectral lines emitted.

The number of different spectral lines equals the number of possible downward transitions among the energy levels up to $n = 4$.

Using the formula:

$$N = \frac{n(n-1)}{2}$$

Substitute $n = 4$:

$$N = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$$

So, the possible transitions are:

$$4 \rightarrow 3,\; 4 \rightarrow 2,\; 4 \rightarrow 1,\; 3 \rightarrow 2,\; 3 \rightarrow 1,\; 2 \rightarrow 1$$

Therefore, the total number of spectral lines is $6$. The correct option is A.

Given: An electron is in the **$4$**th energy level of atomic hydrogen.

Find: Total number of emitted spectral lines.

List all possible downward transitions:

$$n=4 \text{ to } n=3,2,1$$ $$n=3 \text{ to } n=2,1$$ $$n=2 \text{ to } n=1$$

Counting them gives:

$$3 + 2 + 1 = 6$$

Thus, the electron can produce $6$ spectral lines on de-excitation. The correct option is A.