Question

1. What is principal quantum number of the highest energy level for which
the lines in the Balmer series of emission spectra of atomic hydrogen
will just be resolved by a spectrometer of resolving power lambda/d(lambda) = 8 x 1039
(a) 30
(6) 40
(d) 60
(C) 50​

Answer 1

Answer:

d 60

Explanation:

Answer 2

Given:

resolving power = 8 × 10³⁹

To find:

principal quantum number of the highest energy level =?

Step-to-step-explanation:

• The resolving power of spectrum through the Balmer series is given by

         $\displaystyle \frac{1}{\lambda}=R( \frac{1}{n_1^2} -\frac{1}{n_2^2})$

         where

         $\frac{1}{\lambda}$ = resolving power of emission spectra

         R = Rydberg constant

         n₁ = the lower energy level

         n₂ = the higher energy level

• We have to find the principal quantum number of the highest energy level for which the lines in the Balmer series of emission spectra of atomic hydrogen will just be resolved by a spectrometer.

         i.e. n₂ = 1

• Put these values in above equation:

           $\displaystyle 8 \times 10^{39} =1.09737\times 10^7 ( \frac{1}{n_1^2}-\frac{1}{60^2})$

           n₁ = 60

Correct answer is : (d) 60