A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

In ground state, energy of gaseous hydrogen at room temperature = – 13.6 eV.

Energy of electron beam used to bombard the gaseous hydrogen = 12.5 eV

Therefore, after bombardment,
Energy of gaseous hydrogen = –13.6 + 12.6
= – 1.1 eV.

Since,

E_{n} = \frac{- 13.6}{n^{2}}

So, n^{2} = \frac{- 13.6}{- 1.1} = 12.3 \Rightarrow n = 3

That is, the electron would jump from n = 1 to n = 3, where

E_{3} = - \frac{13.6}{3^{2}} = - 1.5 eV .

On de-excitation, the electron may jump from n = 3 to n = 2 giving rise to Balmer series.

It may also jump from n = 3 to n = 1, giving rise to Lyman series.

So, number of spectral line is,

= \frac{n (n - 1)}{2} = \frac{3 (3 - 1)}{2} = 3

Thus, three spectral lines appear.

390 Views

Atoms

Hope you found this question and answer to be good. Find many more questions on Atoms with answers for your assignments and practice.

Physics Part II

Browse through more topics from Physics Part II for questions and snapshot.