Question

the wavelength of spectral line in the lyman series of a H-atom is `1028 Å`. If instead of hydrogen, we consider deuterium then shift in the wavelength of this line be `(m_p=1860m_e)`
A. `1027.7Å`
B. `1036Å`
C. `1028Å`
D. `1021Å`

Answer 1

Correct Answer - A
If `lambda_D` and `lambda_H` are wavelength emitted in the case of deuterium and hydrogen
`therefore (lambda_D)/(lambda_H)=(1-(m_e)/(2m_p))`
Here, `lambda_H=1028Å, m_p=1860m_e`
`therefore lambda_D=(1-(1)/(2xx1860))lambda_H=(1-(1)/(3720))lambda_H`
`lambda _D0.9997lambda_H=0.99973xx1028Å=1027.7Å`.