Question

Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be

• A. One
• B. Two
• C. Three
• D. Four.

Answer 1

Answer: (C)

Three

Answer 2

Ionisation potential of hydrogen atom is $13.6$ eV. Energy required for exciting the hydrogen atom in the ground state to orbit n is given by

$E = E_{n} - E_{1}$

i.e. $12.1 = - \frac{13.6}{n^{2}} - \left( \frac{- 13.6}{1^{2}} \right) = \frac{13.6}{n^{2}} + 13.6$

or, $- 1.5 = \frac{- 13.6}{n^{2}}$ or $n^{2} = \frac{13.6}{1.5} = 9$ or n = 3

number of spectral lines emitted

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