A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be

Question

Match List-I with List-II.

List-I	List-II
(A) Heat capacity of body	(I) \( J\,kg^{-1} \)
(B) Specific heat capacity of body	(II) \( J\,K^{-1} \)
(C) Latent heat	(III) \( J\,kg^{-1}K^{-1} \)
(D) Thermal conductivity	(IV) \( J\,m^{-1}K^{-1}s^{-1} \)

Answer 1

To calculate the number of spectral lines emitted when a 12.5 eV electron beam bombards gaseous hydrogen, we need to consider the energy levels of the hydrogen atom.

In a hydrogen atom, electrons can occupy discrete energy levels given by:

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

where n is the principal quantum number.

Given that the electron beam energy is 12.5 eV, it can excite the hydrogen electron from the ground state (n = 1) to a higher energy level.

The energy difference between two levels is:

\Delta E = E_{n_f} - E_{n_i}

To find how high the electron can go using the given energy of 12.5 eV, we set:

12.5 = -\frac{13.6}{n^2} + (-13.6)

Solving for n:

12.5 = 13.6\left(1 - \frac{1}{n^2}\right)

\frac{12.5}{13.6} = 1 - \frac{1}{n^2}

\frac{1}{n^2} = 1 - \frac{12.5}{13.6}

\frac{1}{n^2} = \frac{1.1}{13.6} \approx 0.0809

n^2 \approx \frac{1}{0.0809} \approx 12.36

n \approx \sqrt{12.36} \approx 3.5

Since n must be an integer, the closest achievable level is n = 3.

The electron is excited to n = 3 and can transition back down to lower energy levels, emitting photons as it does. The possible transitions are:

From n = 3 to n = 2
From n = 3 to n = 1
From n = 2 to n = 1

Thus, a total of 3 spectral lines will be emitted.

The correct answer is: 3.

Answer 2