Question

A series of lines in the spectrum of atomic H lies at wavelengths 656.46 , 486.27 , 434.17 , 410.29 mm. what is the wavelength of next line in this series?

Answer 1

The given series lies in the visible region and thus appears to be Balmer series

As we know, for Balmer series, n₁ = 2

We have to find the at first n₂ for the wavelength 410.29 × 10⁻⁷ cm

λ = 410.29 × 10⁻⁷ cm and n₁ = 2

n₂ can be calculated by this formula:

1/λ = R [ 1/n₁² - 1/n₂² ]

1 / 410.29 × 10⁻⁷ = 109673 [ 1/2² - 1/n₂² ]

Solving this, we get

n₂ = 6

Next line will be obtained during the jump of electron from 7th to 2nd shell, so

1 / λ = 109673 [ 1/2² - 1/7² ]

1 / λ = 109673 [ 1/4 - 1/49 ]

λ = 397.2 × 10⁻⁷ cm

λ = 397.2 nm

Therefore, the wavelength for the next line will be 397.2 nm