Q43 / 50 · Manipal (MET) 2024 · Physics

If an electron in $n=4$ orbit of hydrogen atom jumps down to $n=3$ orbit, the amount of energy released and the wavelength of radiation emitted are

  • Correct answer
    A
    $0.66 \mathrm{eV}, 1.88 \times 10^{-6} \mathrm{~m}$
  • B
    $1.89 \mathrm{eV}, 1.98 \times 10^{-7} \mathrm{~m}$
  • C
    $0.29 \mathrm{eV}, 1.78 \times 10^{-5} \mathrm{~m}$
  • D
    $0.98 \mathrm{eV}, 0.93 \times 10^{-6} \mathrm{~m}$

Solution

$\begin{aligned} E & =\frac{-13.6}{n^2} \mathrm{eV} \\ E_4-E_3 & =13.6\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]\end{aligned}$ $\begin{aligned} & =13.6\left[\frac{7}{144}\right] \mathrm{eV}=0.66 \mathrm{eV} \\ \lambda & =\frac{1}{R\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]}=\frac{144}{1.09 \times 10^7 \times 7} \mathrm{~m} \\ & =1.88 \times 10^{-6} \mathrm{~m}\end{aligned}$

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

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