Question

The wave number of the spectral line in the emission spectrum of hydrogen will be equal to 8 / 9 times the Rydberg’s constant if the electron jumps from

a. n = 3 to n = 1

b. n = 10 to n = 1

c. n = 9 to n = 1

d. n = 2 to n = 1

Answer 1

Correct option is (a) n = 3 to n = 1

 \(\bar{V}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \)

\(\frac89\times R_H=R_H\left(\frac{n_2^2-n_1^2}{n_1^2n_2^2}\right)\)

\(\frac{(n_2-n_1)(n_2+n_1)}{n_1^2n_2^2}=\frac89\)

choosing \(n_1=1,n_2=3\)

then, \(\frac{(3-1)(3+1)}{3^2\times1^2}=\frac{2\times4}{9\times1}=\frac89\)

\(\therefore\) n = 3 to n = 1.

Answer 2

Correct option is a. n = 3 to n = 1

Answer 3

Correct option is a. n = 3 to n = 1

Wave number of spectral line in emission spectrum of hydrogen,

Hence, electron jumps from n₂ =3 to n₁ =1