Question

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

• A. 2 lines in the Lyman series and 1 line in the Balmar series
• B. 3 lines in the Lyman series
• C. 1 line in the Lyman series and 2 lines in the Balmar series
• D. 3 lines in the Balmer series

Answer 1

The Correct Option is A

To solve this problem, we need to understand the energy transitions in the hydrogen atom and how they relate to the emission series like the Lyman and Balmer series. The key here is to determine which electronic transitions are allowed given a particular energy level of excitation.

When a gaseous hydrogen atom is bombarded with a $12.5\, eV$ electron beam, the electrons in the hydrogen atoms are excited to higher energy levels. To understand the possible transitions, let's recall the energy levels of hydrogen atoms, given by the formula:

\(E_n = - \frac{13.6}{n^2}\, eV\), where \(n\) is the principal quantum number.

For hydrogen:

• The ground state energy (\(n=1\)) is \(-13.6\, eV\).
• The first excited state (\(n=2\)) energy is \(-3.4\, eV\).
• The second excited state (\(n=3\)) energy is \(-1.51\, eV\).
• The third excited state (\(n=4\)) energy is \(-0.85\, eV\).

When the electron beam with energy 12.5 eV interacts with hydrogen atoms at the ground state:

The energy of the electron beam is enough to excite the hydrogen atom up to level \(n=3\) because:

• Energy to excite from \(n=1\) to \(n=2\) is \(10.2\, eV\) (from \(-13.6\, eV\) to \(-3.4\, eV\)).
• Energy to excite from \(n=1\) to \(n=3\) is \(12.09\, eV\) (from \(-13.6\, eV\) to \(-1.51\, eV\)).

Therefore, the excitation to \(n=3\) is possible but not to \(n=4\) (requires \(12.75\, eV\)).

From \(n=3\), the electron can transition back down to \(n=1\) and \(n=2\):

• The transition \(n=3 \to n=1\) emits a line in the Lyman series.
• The transition \(n=3 \to n=2\) emits a line in the Balmer series.
• The transition \(n=2 \to n=1\) emits another line in the Lyman series.

Thus, the correct answer is "2 lines in the Lyman series and 1 line in the Balmar series". This matches the provided option.