Q41 / 180 · AIIMS 2017 · Chemistry

Ionisation potential of hydrogen atom is $13.6 \mathrm{eV}$. Hydrogen atom in ground state is excited by monochromatic light of energy $12.1 \mathrm{eV}$. The spectral lines emitted by hydrogen according to Bohr's theory will be

  • A
    one
  • B
    two
  • Correct answer
    C
    three
  • D
    four

Solution

Since ionisation potential of hydrogen atom is $13.6 \mathrm{eV}$. $\therefore \quad E_1=-13.6 \mathrm{eV}$ Now, $E_n-E_1=\frac{-13.6}{n^2}-(-13.6)=12.1$ $\begin{aligned} & \frac{-13.6}{n^2}+13.6=12.1 \\ & \therefore \quad n=3\end{aligned}$ After absorbing $12.1 \mathrm{eV}$, the electron of $\mathrm{H}$-atom is excited to $3^{\text {rd }}$ shell. Thus, possible transitions are 3 i.e., $3 \rightarrow 2$, $2 \rightarrow 1$ and $3 \rightarrow 1$

Hence, the correct answer is option: 'C'.

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