Question 121: Atoms & Nuclei

Question

In hydrogen-like atom $(z=11)$, $n$th line of Lyman series has wavelength A equal to the de Broglie's wavelength of electron in the level from which it originated. What is the value of $n$ ?

Accepted Answer

The $n^{	ext {th }}$ line in Lyman series corresponds to transition $(n+1) ightarrow 1$. The wavelength of this transition is given by
$$ \begin{aligned} \frac{1}{\lambda} & =\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}ight| \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\ \mathrm{R} & =1.097 	imes 10^7 \mathrm{~m}^{-1} 	ext { and } z=11 . \end{aligned} $$
The angular momemtum in $n^{	ext {th }}$ orbit is given by
$$ m v r=\frac{n h}{2 \pi} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
The de-Broglie wavelength of electron in $(n+1)^{	ext {th }}$ orbit is
$$ \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{2 \pi r}{(n+1)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$$
$$ \begin{aligned} &	ext { The radius of }(n+1) 	ext { th orbit is }\ &r=a_0(n+1)^2 / z \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $$
where, $a_0=0.5269 	imes 10^{-10} m$ is Bohr's radius.
Substitute $r$ from equation (iv) into equation (iii) and then substitute into equation (i) to get.
$$ \begin{aligned} \lambda & =\frac{2 \pi\left(a_0(n+1)^2ight) / z}{(n+1)} \ & =\frac{2 \pi a_0(n+1)}{z} \end{aligned} $$
From (i)
$$ \begin{aligned} & \frac{\frac{1}{\frac{2 \pi a_0(n+1)}{z}}=\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}ight|}{\frac{z}{2 \pi a_0(n+1)}=\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}ight|} \ & \frac{z}{2 \pi\left(0.529 	imes 10^{-10}ight)(n+1)} \end{aligned} $$
$$ \begin{aligned} &=1.09 	imes 10^7 	imes z^2\left[1-\frac{1}{(n+1)^2}ight]\ &	ext { Put } z=11\ &\begin{aligned} & \frac{11}{2 \pi\left(0.529 	imes 10^{-10}ight)(n+1)} \ & \quad=1.09 	imes 10^7(11)^2\left[1-\frac{1}{(n+1)^2}ight] \ & \frac{1}{2 \pi\left(0.529 	imes 10^{-10}ight)(n+1)} \ & \quad=1.09 	imes 10^7(10)\left[1-\frac{1}{(n+1)^2}ight] \ & \frac{1}{2 \pi\left(0.529 	imes 10^{-10}ight)\left(1.09 	imes 10^7ight)(11)} \ & \quad=\left[1-\frac{1}{(n+1)^2}ight][n+1] \end{aligned} \end{aligned} $$
$$ \begin{aligned} & \frac{1}{2 \pi(0.529)\left(10^{-10}ight)\left(1.09 	imes 10^7ight)(11)} \ & =\frac{(n+1)^2-1}{(n+1)^2} \cdot(n+1) \end{aligned} $$
$$ \begin{aligned} \frac{1}{2 \pi(0.529)(1.09)(11)\left(10^{-3}ight)} & =\frac{(n+1)^2-1}{(n+1)} \ \frac{1}{39.852 	imes 10^{-3}} & =\frac{(n+1)^2-1}{(n+1)} \ \frac{1000}{39.852} & =\frac{(n+1)^2-1}{(n+1)} \ 25.09 & =\frac{(n+1)^2-1}{n+1} \ \Rightarrow \quad 25 n+25=n^2+2 n & +1-1 \ & \quad 	ext { (approximate) } \end{aligned} $$
$$ \begin{aligned} & \quad \begin{aligned} 25 n & +25=n^2+2 n \ n^2-23 n & -25=0 \ n & =\frac{-(-23) \pm \sqrt{(-23)^2-4.1 .(-25)}}{2.1} \ & =\frac{23 \pm \sqrt{529+100}}{2}=\frac{23 \pm \sqrt{629}}{2} \ n & =\frac{23+\sqrt{629}}{2}=\frac{23-\sqrt{629}}{2} \ n & =\frac{23+25}{2}=\frac{23-25}{2} \  \quad n & =24, n=1 \	herefore n & =24 \end{aligned} \end{aligned} $$

Atoms & Nuclei

In hydrogen-like atom (z=11), nth line of Lyman series has wavelength A equal to the de Broglie's wavelength of electron in the level from which it originated. What is the value of n ?

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