The wavelength of the radiation emitted is \( \lambda_0 \) when an electron jumps from the third to the second orbit of the hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{20}{27}\lambda_0 \)
In the Bohr model the energy of level n is \( E_n=-13.6\,\mathrm{eV}/n^2 \), and a jump from \( n_i \) to \( n_f \) emits a photon of energy \( h\nu=E_i-E_f \). With \( \nu=c/\lambda \) this gives
\( \displaystyle \frac{hc}{\lambda}=13.6\,\mathrm{eV}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \)
so the wavelength is inversely proportional to the bracket.
For the jump from n = 3 to n = 2:
\( \begin{aligned} \frac{hc}{\lambda_0} &= 13.6\,\mathrm{eV}\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ &= 13.6\,\mathrm{eV}\times\frac{5}{36} \end{aligned} \)
For the jump from n = 4 to n = 2:
\( \begin{aligned} \frac{hc}{\lambda} &= 13.6\,\mathrm{eV}\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ &= 13.6\,\mathrm{eV}\times\frac{3}{16} \end{aligned} \)
Dividing the first by the second:
\( \begin{aligned} \frac{\lambda}{\lambda_0} &= \frac{5/36}{3/16} \\ &= \frac{80}{108} \\ &= \frac{20}{27} \end{aligned} \)
\( \displaystyle \Rightarrow\; \lambda=\frac{20}{27}\lambda_0 \)