Question

Caution: this is tricky, try doing it using the modified Rydberg's formula. A particular H-like ion emits radiation of frequency $2.467\times {10}^{15}$ Hz when it makes a transition from n = 2 to n = 1. What will be the frequency of radiation in a transition from n = 3 to n = 1? Think carefully about why the first part of the question is even needed (it is needed).

• A. Hz
• B. Hz
• C. Hz
• D. Hz

Answer 1

The correct option is A Hz

So its given that frequency $v_{2-1}$ = $2.467\times {10}^{15}$ Hz
$\lambda v=C$
$\frac{1}{\lambda }=\frac{v}{c}$ ......(i)
Since in the question its mentioned hydrogen like atom.
So we will have to use the modified Rydberg’s formula
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ ....(ii)

Transition is happening from 2 to 1
$\Rightarrow n_2=2;n_1=1$
From (i) and (ii)
$\frac{v_2-1}{C}=R{z}^2(\frac{1}{1^2}-\frac{1}{2^2})$
$\Rightarrow \frac{v_2-1}{C}=R{z}^2\left(\frac{3}{4}\right)$ ....(iii)
In the next part, trasition is happening from n = 3 to n = 1
So $n_2$ = 3 to $n_1$ = 1
$\Rightarrow \frac{v_3-1}{C}=R{z}^2\left(\frac{8}{9}\right)$ ....(iv)
Dividing (iii) and (iv) we get
$\frac{v_{2-1}}{v_{3-1}}=\frac{3\times 9}{4\times 8}$
$v_{3-1}=\frac{32\times 2.29\times {10}^{15}}{27}$
= 2.71 $\times$ ${10}^{15}$ Hz