A hydrogen atom in its ground state absorbs a photon. After this, the radius of the electron orbit is 9 times the ground-state radius. What is the frequency of the photon? (Take \(E_n=-13.6/n^2\ \mathrm{eV}\), \(h=6.6\times10^{-34}\ \mathrm{J\,s}\) and \(1\ \mathrm{eV}=1.6\times10^{-19}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2.9\times10^{15}\ \mathrm{Hz}\)
In Bohr’s model, the radius of an orbit grows as \( n^2 \). So a radius 9 times the ground-state value means the electron is now in the \( n=3 \) level. The photon energy is the gap \( E_3-E_1 \), and its frequency is this energy divided by \( h \).
Level reached:
\( \begin{aligned} \frac{r_n}{r_1} &= n^2 \\ 9 &= n^2 \\ n &= 3 \end{aligned} \)
Energies of the two levels:
\( E_1=-13.6\ \mathrm{eV} \)
\( \begin{aligned} E_3 &= -\frac{13.6}{3^2} \\ &= -1.51\ \mathrm{eV} \end{aligned} \)
Energy of the photon:
\( \begin{aligned} E &= E_3-E_1 \\ &= -1.51-(-13.6) \\ &= 12.09\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} E &= 12.09\times1.6\times10^{-19} \\ &= 1.93\times10^{-18}\ \mathrm{J} \end{aligned} \)
Frequency of the photon:
\( \begin{aligned} \nu &= \frac{E}{h} \\ &= \frac{1.93\times10^{-18}}{6.6\times10^{-34}} \\ &= 2.9\times10^{15}\ \mathrm{Hz} \end{aligned} \)
From the above, the electron is lifted from \( n=1 \) to \( n=3 \). Hence, the photon has a frequency of about \( 2.9\times10^{15}\ \mathrm{Hz} \).
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “The radius of nth possible orbit thus found is”
\( \displaystyle r_n=\frac{n^2}{m}\left(\frac{h}{2\pi}\right)^2\frac{4\pi\varepsilon_0}{e^2} \)\( \qquad (12.7) \)
Energy levels – “… Similarly, E3 = –1.51 eV and E3 – E1 = 12.09 eV, or to excite the hydrogen atom from its ground state (n = 1) to second excited state (n = 3), 12.09 eV energy is required, and so on.”