JEE Physics 2025 Question with Solution

Given: The electron is initially in the fourth excited state, so the initial principal quantum number is $n_i = 5$. The emitted photon energy is $2.86 \, \text{eV}$.

Find: The final principal quantum number $n$.

The energy of an electron in the $k^{th}$ energy level of hydrogen atom is

$$E_k = -\frac{13.6}{k^2} \, \text{eV}$$

Since the electron is initially in the fourth excited state,

$$k_i = 4 + 1 = 5$$

So, the initial energy is

$$E_i = E_5 = -\frac{13.6}{5^2} \, \text{eV} = -\frac{13.6}{25} \, \text{eV}$$ $$E_i = -0.544 \, \text{eV}$$

For emission of a photon,

$$\Delta E = E_i - E_n$$

Given $\Delta E = 2.86 \, \text{eV}$, therefore

$$E_n = E_i - \Delta E$$ $$E_n = -0.544 \, \text{eV} - 2.86 \, \text{eV}$$ $$E_n = -3.404 \, \text{eV}$$

Now use

$$E_n = -\frac{13.6}{n^2} \, \text{eV}$$

So,

$$-3.404 = -\frac{13.6}{n^2}$$ $$n^2 = \frac{13.6}{3.404}$$ $$n^2 \approx 3.995 \approx 4$$ $$n = \sqrt{4} = 2$$

Therefore, the integer value of $n$ is $2$.

Given: Initial state is the fourth excited state, so $n_i = 5$. Photon energy emitted is $2.86 \, \text{eV}$.

Find: Final state $n$.

Use the direct hydrogen transition relation:

$$E_{\text{photon}} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \, \text{eV}$$

Substitute $n_i = 5$ and $E_{\text{photon}} = 2.86 \, \text{eV}$:

$$2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{25} \right)$$

Divide by $13.6$:

$$\frac{2.86}{13.6} = \frac{1}{n^2} - \frac{1}{25}$$ $$0.21029 \approx \frac{1}{n^2} - 0.04$$ $$\frac{1}{n^2} = 0.25029 \approx 0.25 = \frac{1}{4}$$

Hence,

$$n^2 = 4$$ $$n = 2$$

This shortcut works because the photon energy is directly the difference of hydrogen energy levels. Therefore, the final state is $n = 2$.