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JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2018
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The de-Broglie wavelength \(({\lambda _B})\)  associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground  state  \(({\lambda _G})\) by

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Answer & Solution

Correct Answer

(D) \({\lambda _B}\, = \,{3\lambda _{G}}\)

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Detailed explanation

de-Broglie wavelength, \(\lambda=\frac{h}{P}\)

\(\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{G}}}=\frac{\mathrm{P}_{\mathrm{G}}}{\mathrm{P}_{\mathrm{B}}}=\frac{\mathrm{mv}_{\mathrm{G}}}{\mathrm{mv}_{\mathrm{B}}}\)

Speed of electron \(v \propto \frac{z}{n}\)

so \(\frac{\lambda_{B}}{\lambda_{G}}=\frac{n_{B}}{n_{G}}=\frac{3}{1} \Rightarrow \lambda_{B}=3 \lambda_{G}\)

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