JEE Mains · Physics · STD 12 - 12. atoms
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
- A \({\lambda _B}\, = \,{\lambda _{G}}/3\)
- B \({\lambda _B}\, = \,{\lambda _{G}}/2\)
- C \({\lambda _B}\, = \,{2\lambda _{G}}\)
- D \({\lambda _B}\, = \,{3\lambda _{G}}\)
Answer & Solution
Correct Answer
(D) \({\lambda _B}\, = \,{3\lambda _{G}}\)
Step-by-step Solution
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…
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