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de-Broglie wavelength of electron in 2^(...

de-Broglie wavelength of electron in `2^("nd")` excited state of hydrogen atom is: [where `r_(0)` is the radius of `1^("st")` orbit in H-atom]

A

`r_(0)`

B

`pir_(0)`

C

`3pir_(0)`

D

`6pir_(0)`

Text Solution

Verified by Experts

The correct Answer is:
D
`rprop (n^(2))/(z)" "rArr" "r_(o)=kxx(1^(2))/(1)" "rArr" "k=r_(o), r_(3)=kxx(3^(2))/(1)=9k=9r_(o)(2^("nd")" excited state means n = 3")`
If de-Broglie wavelength is `lambda` then `3lambda=2pir_(3)=18pir_(o) rArr ambda=6pir_(o)`
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