A hydrogen atom emits uv radiation of wavelength 102.5 nm. The quantum numbers of the states involved are

3, 1

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2, 1

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4, 1

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4, 2

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Solution

The correct option is A 3, 1
$λ = 102.5 n m$
$Δ E = \frac{1240}{102.5} = 12.09 e V$
which is roughly equal to ${\frac{8}{9}}^{t h} o f 13.6 e V$
Therefore, $Δ E = \frac{h C}{λ} = R_{H} [\frac{8}{9}]$
$Δ E = R_{H} [\frac{1}{R} - \frac{1}{3^{2}}]$
$∴ n_{1} = 1$ and $n_{2} = 3$

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102.5 n m

A hydrogen atom emits ultraviolet radition of wavelen gth 102.5 nm. What are the quantum numbers of the states involved in the transition ?

A hydrogen atom in a state having a binding energy of 0.85 e V makes transition to a state with excitation energy 10.2 e V.

(a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition.

(b) Find the wavelength of the emitted radiation.

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