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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` |
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Answer» 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`. |
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