Question 19
If the wavelength of the first member of Lyman series of hydrogen is $$\lambda$$. The wavelength of the second member will be
Solution
Lyman series: $$\frac{1}{\lambda} = R(1 - 1/n^2)$$.
First member (n=2): $$\frac{1}{\lambda} = R(1-1/4) = \frac{3R}{4}$$.
Second member (n=3): $$\frac{1}{\lambda_2} = R(1-1/9) = \frac{8R}{9}$$.
$$\frac{\lambda_2}{\lambda} = \frac{3R/4}{8R/9} = \frac{27}{32}$$.
$$\lambda_2 = \frac{27}{32}\lambda$$.
The answer is Option (1): $$\frac{27}{32}\lambda$$.
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