Question: Q. 8. The energy levels of a hypothetical atom are given below. Which of the shown transitions will result in the emission of photon of wavelength $275 \mathrm{~nm}$ ?
A] [OD South 2016]
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Solution:
Ans. (i) Energy of photon $=\frac{h c}{\lambda}$
$$ \begin{aligned} & =\frac{6.64 \times 10^{-34} \times 3 \times 10^{8}}{275 \times 10^{-9} \times 1.6 \times 10^{-18}} \mathrm{eV} \ & =4.5 \mathrm{eV} \quad 1 / 2+1 / 2 \end{aligned} $$
(ii) The corresponding transition is $B$
[CBSE Marking Scheme 2016]
Detailed Answer :
Energy of a photon corresponding to wavelength $\lambda$,
$$ \begin{aligned} E & =h \frac{c}{\lambda} \ & =\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{275 \times 10^{-9}} \mathrm{~J} \quad 1 / 2 \ & =\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{275 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV} \end{aligned} $$
$$ \begin{align*} & =\frac{6.6 \times 3 \times 10^{2}}{275 \times 1.6} \ & =4.5 \mathrm{eV} \tag{1} \end{align*} $$
(ii) the calculated energy of the photon matches with the transition $B$.
