Question

Show Work.

A series of lines in the spectrum of atomic hydrogen lies at 656 nm, 486 nm, 434 nm, and 410 nm. What is the wavelength of the next line of the series?

a) 200 nm b) 397 nm c) 410 nm d) 420 nm e) 710 nm

Answer 1

Given that series of lines in the spectrum of atomic hydrogen lies at 656 nm, 486 nm, 434 nm, and 410 nm.

The given series lies in the visible region and fall in Balmer series.

We know that for Balmer series n1 = 2   n2 = 2,3,4------

Lut us take wavelength = 410 nm

Wavelength λ = 410 nm = 410 ×10⁻⁷ cm              [ 1nm = 10^-9 m = 10^-7 cm]

n2 = ?

We know that

1/λ = R_H [1/n1^2−1/n2^2]               R_H = Rydberg constant for Hydrogen atom = 109678 cm^-1

1/ 410 ×10⁻⁷ cm   = 109678 cm^-1 [1/n1^2−1/n2^2]

1/ 410 ×10⁻⁷ cm   = 109678 cm^-1 [1/2^2−1/n2^2]

On simplification,

n2 = 6

Hence,

Next line will be obtained during the transition of electron from 7^th to 2^nd shell.

So,

n1= 2 , n2 =7

1/λ = R_H[1/2^2−1/7^2]

     = 109678 cm^-1 [1/4−1/49]

λ = 397 ×10⁻⁷ cm

λ = 397 nm              [ 1 nm = 10^-7 cm]

Therefore,

answer is (b)