Question

In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series, to the longest wavelength in the Balmer series is,

• A. $\frac{5}{27}$
• B. $\frac{25}{36}$
• C. $\frac{4}{9}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $\frac{5}{27}$

As we know.
$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For the longest wavelength of lyman series, $n_1=1$ and $n_2=2$.

Similarely for longest wavelength in Balmer series $n_1=2$ and $n_2=3$
So, basically we can say that
$({\lambda }_1{)}_{Lyman}={\lambda }_{1max}$ and $({\lambda }_1{)}_{Balmer}={\lambda }_{2max}$

$\frac{1}{{\lambda }_{1max}}=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$=R(\frac{1}{1^2}-\frac{1}{2^2})$

$\Rightarrow {\lambda }_{1max}=\frac{4}{3R}$

And, $\frac{1}{{\lambda }_{2max}}=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$=R(\frac{1}{2^2}-\frac{1}{3^2})$
$\Rightarrow {\lambda }_{2max}=\frac{36}{5R}$

$\Rightarrow \frac{{\lambda }_{1max}}{{\lambda }_{2max}}=\frac{4}{3R}\times \frac{5R}{36}=\frac{5}{27}$

Hence, $\left(A\right)$ is the correct answer.