In a hydrogen atom, the electron falls from the \(n=4\) level to the \(n=2\) level. What is the frequency of the photon emitted? (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 2
The correct answer is Option (2) → \(6.2\times10^{14}\ \mathrm{Hz}\)
The photon carries away the energy difference of the two levels. Its frequency is this energy divided by Planck’s constant, \( \nu=(E_i-E_f)/h \).
Energies of the two levels:
\( \begin{aligned} E_4 &= -\frac{13.6}{4^2} \\ &= -0.85\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} E_2 &= -\frac{13.6}{2^2} \\ &= -3.40\ \mathrm{eV} \end{aligned} \)
Energy of the photon:
\( \begin{aligned} E &= E_4-E_2 \\ &= -0.85-(-3.40) \\ &= 2.55\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} E &= 2.55\times1.6\times10^{-19} \\ &= 4.08\times10^{-19}\ \mathrm{J} \end{aligned} \)
Frequency of the photon:
\( \begin{aligned} \nu &= \frac{E}{h} \\ &= \frac{4.08\times10^{-19}}{6.6\times10^{-34}} \\ &= 6.2\times10^{14}\ \mathrm{Hz} \end{aligned} \)
Therefore, the photon emitted in the jump from \( n=4 \) to \( n=2 \) has a frequency of about \( 6.2\times10^{14}\ \mathrm{Hz} \).
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “Atomic energies are often expressed in electron volts (eV) rather than joules. Since 1 eV = 1.6 × 10–19 J, … can be rewritten as”
\( \displaystyle E_n=-\frac{13.6\ \mathrm{eV}}{n^{2}} \)\( \qquad (12.10) \)
Bohr Model of the Hydrogen Atom – “… It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by”
\( \displaystyle h\nu=E_i-E_f \)\( \qquad (12.6) \)