A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. It will emit:

An electron beam with energy 12.5 eV is used to bombard gaseous hydrogen atoms. At room temperature, hydrogen atoms are in their ground state (n=1). The energy required to excite an electron from the ground state to a higher energy level (n) is given by the formula:

$$\Delta E = E_n - E_1 = -\frac{13.6}{n^2} - \left(-\frac{13.6}{1^2}\right) = 13.6 \left(1 - \frac{1}{n^2}\right) \text{ eV}$$

The bombarding electron has 12.5 eV of energy. We need to find the maximum n such that the excitation energy is less than or equal to 12.5 eV. So, we solve:

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

Rearranging the inequality:

$$13.6 - \frac{13.6}{n^2} \leq 12.5$$

Subtracting 13.6 from both sides:

$$-\frac{13.6}{n^2} \leq 12.5 - 13.6$$

$$-\frac{13.6}{n^2} \leq -1.1$$

Multiplying both sides by -1 (which reverses the inequality):

$$\frac{13.6}{n^2} \geq 1.1$$

Dividing both sides by 13.6:

$$\frac{1}{n^2} \geq \frac{1.1}{13.6}$$

Calculating the right side:

$$\frac{1.1}{13.6} = \frac{11}{136} = \frac{11 \div 2}{136 \div 2} = \frac{5.5}{68} \approx 0.080882$$

So:

$$\frac{1}{n^2} \geq 0.080882$$

Taking reciprocals (and reversing the inequality since both sides are positive):

$$n^2 \leq \frac{1}{0.080882} \approx 12.36$$

Taking the square root:

$$n \leq \sqrt{12.36} \approx 3.516$$

Since n must be an integer, the possible values are n = 2 and n = 3.

Now, verify the energy required for each:

For n = 2:

$$\Delta E = 13.6 \left(1 - \frac{1}{2^2}\right) = 13.6 \left(1 - \frac{1}{4}\right) = 13.6 \times \frac{3}{4} = \frac{40.8}{4} = 10.2 \text{ eV} \leq 12.5 \text{ eV}$$

For n = 3:

$$\Delta E = 13.6 \left(1 - \frac{1}{3^2}\right) = 13.6 \left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} = \frac{108.8}{9} \approx 12.0889 \text{ eV} \leq 12.5 \text{ eV}$$

For n = 4:

$$\Delta E = 13.6 \left(1 - \frac{1}{4^2}\right) = 13.6 \left(1 - \frac{1}{16}\right) = 13.6 \times \frac{15}{16} = \frac{204}{16} = 12.75 \text{ eV} > 12.5 \text{ eV}$$

Since 12.75 eV exceeds 12.5 eV, excitation to n = 4 is not possible. Thus, electrons can only be excited to n = 2 and n = 3.

When these excited electrons fall back to lower energy levels, they emit photons corresponding to spectral lines. The possible transitions are:

• From n = 3 to n = 1 (Lyman series)
• From n = 3 to n = 2 (Balmer series)
• From n = 2 to n = 1 (Lyman series)

This gives two transitions in the Lyman series (n=3→n=1 and n=2→n=1) and one transition in the Balmer series (n=3→n=2).

Now, comparing with the options:

A. 2 lines in the Lyman series and 1 line in the Balmer series

B. 3 lines in the Lyman series

C. 1 line in the Lyman series and 2 lines in the Balmer series

D. 3 lines in the Balmer series

Option A matches the transitions we have identified.

Hence, the correct answer is Option A.