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The radius of the orbit of an electron i...

The radius of the orbit of an electron in a Hydrogen-like atom is `4.5 a_(0)`, where `a_(0)` is the Bohr radius. Its orbital angular momemtum is `(3h)/(2pi)`. It is given that `h` is Plank constant and `R` si Rydberg constant. The possible wavelength `(s)`, when the atom-de-excites. is (are):

A

`(9)/(32R)`

B

`(9)/(16R)`

C

`(9)/(5R)`

D

`(4)/(3R)`

Text Solution

Verified by Experts

The correct Answer is:
A, C
Angular momentum `=(nh)/(2pi)=(3h)/(2pi)implies n=3`
Also `r=n^(2)/2 a_(0) implies (9a_(0))/2=3^(2)/Z a_(0) implies Z=2`
For de-excitation
`1/lambda=Rz^(2) [1/n_(1)^(2)-1/n_(2)^(2)]=4R[1/n_(1)^(2)-1/n_(2)^(2)]`
For `n=3` to `n=1` :
`1/lambda=4R[1/1-1/9] implies lambda=9/(32R)`
For `n=3` to `n=2`:
`1/lambda=4R[1/4-1/9]implies lambda=9/(5R)`
For `n=2` to `n=1` :
`1/lambda=4R[1/1-1/4]implies lambda=3/R`
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