Atoms & Nuclei

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111 of 239

Question 111 of 239

If the wavelength of the first member of Lyman series of hydrogen is λ\lambda. The wavelength of the second member will be

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Answer: (A) 2732λ\dfrac{27}{32}\lambda

For Lyman series:
For first member, 1λ=13.6Z2hc[112−122]\dfrac{1}{\lambda }=\dfrac{13.6{Z}^{2}}{hc}\left[\dfrac{1}{{1}^{2}}-\dfrac{1}{{2}^{2}}\right] ….. (i)
For second member, 1λ′=13.6Z2hc[112−132]\dfrac{1}{\lambda '}=\dfrac{13.6{Z}^{2}}{hc}\left[\dfrac{1}{{1}^{2}}-\dfrac{1}{{3}^{2}}\right] ….. (ii)
On dividing (i) and (ii), we get
λ′=2732λ\lambda '=\dfrac{27}{32}\lambda

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