Question

In which transition of a hydrogen atom ,photons of lowest frequency are
emitted?
1)n=4 to n=3
2)n=4 to n=2
3)n=2 to n=1
4)n=3 to n=1

Answer 1

Dear Student,
The formula for frequency of Photon due to transition of electron from one energy state of hydrogen atom to the other is given by-
$$\begin{gathered} FrequencyofPhotonemitted=\nu =Rc\left[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right] \\ HereR=Rydbergcons\tan tandc=velocityoflight. \\ {n}_1=Initialstateand{n}_2=finalstate \\ When{n}_1=3and{n}_2=4 \\ Frequency=\nu =Rc\left[\frac{1}{3^2}-\frac{1}{4^2}\right]=0.049Rc \\ When{n}_1=2and{n}_2=4 \\ Frequency=\nu =Rc\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=0.1875Rc \\ When{n}_1=1and{n}_2=2 \\ Frequency=\nu =Rc\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=0.75Rc \\ When{n}_1=1and{n}_2=3 \\ Frequency=\nu =Rc\left[\frac{1}{1^2}-\frac{1}{3^2}\right]=9Rc \end{gathered}$$
Thus the frequency of photon is minimum when there is a transition from n = 4 to n=3.
Regards.