Question

If the first line in the Lyman series has wavelength $\lambda$, then the first line in Balmer series has the wavelength

• A. $\frac{27}{5}\lambda$
• B. $\frac{32}{27}\lambda$
• C. $\frac{28}{21}\lambda$
• D. $\frac{15}{4}\lambda$

Answer 1

Answer: (A)

Wavelength for Lyman series is calculated as
$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$
For first line of Lyman series, $n=2$
$\therefore \frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
$\Rightarrow \frac{1}{\lambda }=\frac{3R}{4}$
$\Rightarrow \lambda =\frac{4}{3R}$
$\Rightarrow R=\frac{4}{3\lambda }$
Wavelength of Balmer series is calculated as
$\frac{1}{{\lambda }^{\prime }}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$
For first line of Balmer series, $n=3$
$\therefore \frac{1}{{\lambda }^{\prime }}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$
$\Rightarrow \frac{1}{{\lambda }^{\prime }}=\frac{5}{36}R$
$\Rightarrow \frac{1}{{\lambda }^{\prime }}=\frac{5}{36}\times \frac{4}{3\lambda }$ [from Eq. (i)]
$\Rightarrow \frac{1}{{\lambda }^{\prime }}=\frac{5}{27\lambda }$
$\Rightarrow {\lambda }^{\prime }=\frac{27\lambda }{5}$