Question

A hydrogen atom in its fourth excited state emits a photon with a wavelength of $1282$nm. What is the atom’s maximum possible orbital angular momentum (as a multiple of $h$) after the emission?

Short answer

Maximum possible orbital angular momentum is $\frac{3h}{2\mathrm{\pi }}$.

Step-by-step solution

:  Given information

Fourth excited state means $n=5$

Wavelength, $\lambda =1282nm=1282\times {10}^{-9}m$

:  Simplification

Using Formula , $\frac{1}{\lambda }=R{z}^2\left(\frac{1}{x^2}-\frac{1}{n^2}\right)$, where $\lambda$is the wavelength , $R$is rydberg constant , $x$is the final orbital of electron , $n$is the initial orbital of electron , $z$is atomic no

Now for hydrogen $z=1$

Now , Putting all the values in the formula we get $\frac{1}{1282\times {10}^{-9}}=1.01\times {10}^7\left(\frac{1}{x^2}-\frac{1}{5^2}\right)$

Solving it we get $x=3$

It means $n=5\Rightarrow n=3$transition

Now , we know that orbital angular momentum is $\frac{nh}{2\mathrm{\pi }}$

Now after emmision we have$n=3$

So, the maximum angular momentum comes out to be $\frac{3h}{2\mathrm{\pi }}$