Question

If the wavelength of the first member of the Lyman series of hydrogen is $\lambda$. The wavelength of the second member will be:

• A. $\frac{27}{32} \lambda$
• B. $\frac{32}{27} \lambda$
• C. $\frac{27}{5} \lambda$
• D. $\frac{5}{27} \lambda$

Answer 1

Answer: (A)

$\frac{27}{32} \lambda$

Answer 2

For the first member of the Lyman series:

$\frac{1}{\lambda} = \frac{13.6 \, z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] \quad \dots \text{(i)}$

For the second member of the Lyman series:

$\frac{1}{\lambda'} = \frac{13.6 \, z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] \quad \dots \text{(ii)}$

Dividing equation (i) by (ii):

$\lambda' = \frac{27}{32} \lambda$