Question

Ratio of n${}^{th}$ to m${}^{th}$ wavelength of Lyman series in $H$ - atom is equal to:

• A. $\frac{{\lambda }_n}{{\lambda }_m}={\left\{\frac{(m+1)}{(n+1)}\right\}}^2\left\{\frac{(n+1{)}^2-1}{(m+1{)}^2-1}\right\}$
• B. $\frac{{\lambda }_n}{{\lambda }_m}=\frac{(m^2-1)\times n^2}{(n^2-1)\times m^2}$
• C. $\frac{{\lambda }_n}{{\lambda }_m}=\frac{(n^2-1)\times m^2}{(m^2-1)\times n^2}$
• D. $\frac{{\lambda }_n}{{\lambda }_m}={\left\{\frac{(n+1)}{(m+1)}\right\}}^2\left\{\frac{(m+1{)}^2-1}{(n+1{)}^2-1}\right\}$

Answer 1

The correct option is D $\frac{{\lambda }_n}{{\lambda }_m}={\left\{\frac{(n+1)}{(m+1)}\right\}}^2\left\{\frac{(m+1{)}^2-1}{(n+1{)}^2-1}\right\}$

According to Bohr model,

$\frac{1}{{\lambda }_m}=R_H[\frac{1}{1^2}-\frac{1}{(m+1{)}^2}]m$th line from $m+1$ to $1$

$\frac{1}{{\lambda }_m}=R_H\left[\frac{(m+1{)}^2-1}{(m+1{)}^2}\right]$

$\frac{1}{{\lambda }_n}=R_H[\frac{1}{1^2}-\frac{1}{(n+1{)}^2}]n$th line from $n+1$ to $1$

$\frac{1}{{\lambda }_n}=R_H\left[\frac{(n+1{)}^2-1}{(n+1{)}^2}\right]$

Now, ratio:

$\therefore \frac{{\lambda }_n}{{\lambda }_m}=\left\{\frac{(m+1{)}^2-1}{(n+1{)}^2-1}\right\}\times \frac{(n+1{)}^2}{(m+1{)}^2}$