Question

The ionisation potential of hydrogen atom is 13.6 eV. The hydrogen atom in the ground state is excited by the monochromatic light of energy 12.1 eV.

The spectral lines emitted by hydrogen according to Bohr's theory will be:

• A. one
• B. two
• C. three
• D. four

Answer 1

Answer: (C)

three

Since ionisation potential of hydrogen atom is 13.6 eV.

$\therefore E_1=-13.6eV$

Now, $E_n-E_1=\frac{-13.6}{n^2}-(-13.6)=12.1$

$\frac{-13.6}{n^2}+13.6=12.1$

$\therefore n=3$

After absorbing 12.1 eV, the electron of H-atom is excited to $3^{rd}$ shell.

Thus, possible transactions are 3 i.e $3\to 2,2\to 1$ and $3\to 1$