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Q.

The energy level of an atom is shown in the figure. Which one of the transitions will result in the emission of photon of wavelength 124.1nm. Given $h = 6.62 \times 10^{- 34} Js$ .

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a

B

b

A

c

C

d

D

answer is D.

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Detailed Solution

$\begin{aligned} {ΔE}_{A} = 2.2 eV \\ {ΔE}_{B} = 5.2 eV \\ {ΔE}_{C} = 3 eV \\ {ΔE}_{D} = 10 eV \end{aligned}$

$λ = \frac{hc}{ΔE}$

$\begin{array}{l} λ_{A} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2.2 \times 1.6 \times 10^{- 19}} \\ = \frac{12.41 \times 10^{- 7}}{2.2} m \\ = \frac{1241}{2.2} nm = 564 nm \\ λ_{B} = \frac{1241}{5.2} nm = 238.65 nm \\ λ_{C} = \frac{1241}{3} nm = 413.66 nm \\ λ_{D} = \frac{1241}{10} = 124.1 nm \end{array}$

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