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QJEEM24Ph60d1: JEE Main 2024 Physics - Class 12 JEEM24Ph60d1 Solution: If the wavelength of the first member of Lyman series of…

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

Multiple Choice Options

  • A 275λ\frac{27}{5} \lambda
  • B 527λ\frac{5}{27} \lambda
  • C 2732λ\frac{27}{32} \lambda
  • D 3227λ\frac{32}{27} \lambda

Step-by-Step Verified Solution

amp;1λ=13.6z2hc[112−122]......(i)amp;1λ′=13.6z2hc[112−132]......(ii)\begin{aligned}& \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] ...... \text{(i)}\\& \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] ...... \text{(ii)}\end{aligned}

On dividing (i) & (ii)

λ′=2732λ\lambda^{\prime}=\frac{27}{32} \lambda

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