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Question

The ratio of the frequencies of the long wavelength limits of the Balmer and Lyman series of hydrogen is


A
27:5
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B
5:27
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C
4:1
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D
1:4
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Solution

The correct option is D $$27:5$$
Longest wavelengths are formed when the electron falls from next highest orbit to the orbit of the series releasing the minimum amount of energy.
$$E(Balmer) = K\left [ 1 - \dfrac{1}{4} \right ] = \dfrac{3k}{4}$$ --------- (I)
$$E(Lyman) = K\left [ \dfrac{1}{4} - \dfrac{1}{9} \right ] = \dfrac{5K}{36}$$----------(II)
Ratio of (I) and (II),
Balmer : Lyman
$$=\dfrac{3K}{4} \times \dfrac{36}{5K}$$

$$= 27:5$$

Physics

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