The $n^{\text {th }}$ line in Lyman series corresponds to transition $(n+1) \rightarrow 1$. The wavelength of this transition is given by

$$ \begin{aligned} \frac{1}{\lambda} & =\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right| \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \mathrm{R} & =1.097 \times 10^7 \mathrm{~m}^{-1} \text { and } z=11 . \end{aligned} $$

The angular momemtum in $n^{\text {th }}$ orbit is given by

$$ m v r=\frac{n h}{2 \pi} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$

The de-Broglie wavelength of electron in $(n+1)^{\text {th }}$ orbit is

$$ \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{2 \pi r}{(n+1)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$$

$$ \begin{aligned} &\text { The radius of }(n+1) \text { th orbit is }\\ &r=a_0(n+1)^2 / z \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $$

where, $a_0=0.5269 \times 10^{-10} m$ is Bohr's radius.

Substitute $r$ from equation (iv) into equation (iii) and then substitute into equation (i) to get.

$$ \begin{aligned} \lambda & =\frac{2 \pi\left(a_0(n+1)^2\right) / z}{(n+1)} \\ & =\frac{2 \pi a_0(n+1)}{z} \end{aligned} $$

From (i)

$$ \begin{aligned} & \frac{\frac{1}{\frac{2 \pi a_0(n+1)}{z}}=\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right|}{\frac{z}{2 \pi a_0(n+1)}=\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right|} \\ & \frac{z}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \end{aligned} $$

$$ \begin{aligned} &=1.09 \times 10^7 \times z^2\left[1-\frac{1}{(n+1)^2}\right]\\ &\text { Put } z=11\\ &\begin{aligned} & \frac{11}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \\ & \quad=1.09 \times 10^7(11)^2\left[1-\frac{1}{(n+1)^2}\right] \\ & \frac{1}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \\ & \quad=1.09 \times 10^7(10)\left[1-\frac{1}{(n+1)^2}\right] \\ & \frac{1}{2 \pi\left(0.529 \times 10^{-10}\right)\left(1.09 \times 10^7\right)(11)} \\ & \quad=\left[1-\frac{1}{(n+1)^2}\right][n+1] \end{aligned} \end{aligned} $$

$$ \begin{aligned} & \frac{1}{2 \pi(0.529)\left(10^{-10}\right)\left(1.09 \times 10^7\right)(11)} \\ & =\frac{(n+1)^2-1}{(n+1)^2} \cdot(n+1) \end{aligned} $$

$$ \begin{aligned} \frac{1}{2 \pi(0.529)(1.09)(11)\left(10^{-3}\right)} & =\frac{(n+1)^2-1}{(n+1)} \\ \frac{1}{39.852 \times 10^{-3}} & =\frac{(n+1)^2-1}{(n+1)} \\ \frac{1000}{39.852} & =\frac{(n+1)^2-1}{(n+1)} \\ 25.09 & =\frac{(n+1)^2-1}{n+1} \\ \Rightarrow \quad 25 n+25=n^2+2 n & +1-1 \\ & \quad \text { (approximate) } \end{aligned} $$

$$ \begin{aligned} & \quad \begin{aligned} 25 n & +25=n^2+2 n \\ n^2-23 n & -25=0 \\ n & =\frac{-(-23) \pm \sqrt{(-23)^2-4.1 .(-25)}}{2.1} \\ & =\frac{23 \pm \sqrt{529+100}}{2}=\frac{23 \pm \sqrt{629}}{2} \\ n & =\frac{23+\sqrt{629}}{2}=\frac{23-\sqrt{629}}{2} \\ n & =\frac{23+25}{2}=\frac{23-25}{2} \\ \quad n & =24, n=1 \\\therefore n & =24 \end{aligned} \end{aligned} $$

JEE Advanced Subjects