Hydrogen atom in its ground state is excited by a radiation of wavelength 97.26 nm. The longest wavelength it emits is

A
1250 nm
B
1400 nm
C
1875 nm
D
2050 nm

Solution:

$Δ E = \frac{h c}{λ} = \frac{(6.626 \times 10^{- 34} Js) (3 \times 10^{8} {ms}^{- 1})}{(97.26 \times 10^{- 9} m)} = 2.0438 \times 10^{- 18} J$

Also $Δ E = (2.18 \times 10^{- 18} J) (\frac{1}{1^{2}} - \frac{1}{n_{2}^{2}})$

Hence $\begin{array}{l} \frac{1}{n_{2}^{2}} = 1 - (\frac{Δ E}{2.18 \times 10^{- 18} J}) = 1 - \frac{2.0436 \times 10^{- 18}}{2.18 \times 10^{- 18}} = 1 - 0.9374 \\ n_{2} = \sqrt{1 / (1 - 0.9374)} ≃ 4 \end{array}$

The transition $n_{2} = 4 \to n_{1} = 3$ will emit the longest wavelength. Hence

$\begin{array}{l} Δ E = (2.18 \times 10^{- 18} J) (\frac{1}{3^{2}} - \frac{1}{4^{2}}) = 1.06 \times 10^{- 19} J \\ λ = \frac{h c}{Δ E} = \frac{(6.626 \times 10^{- 34} Js) (3 \times 10^{8} {ms}^{- 1})}{(1.06 \times 10^{- 19} J)} = 1.875 \times 10^{- 6} m = 1875 nm \end{array}$

Solution:

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