The radius of the orbit of an electron in a Hydrogen - like atom is `4.5s_(0)` where `a_(0)` is the bohr radius its orbital angular momentum is `(3h)/(2 pi) ` it is given that is is plank constant and R is rydberg constant .The possible wavelength `(s)` , when the atom de- excite , is (are)

Options

A. `9/(32R)`
B. `9/(16R)`
C. `9/(5R)`
D. `4/(3R)`

Answer

A, C

Explanation

As `mvr=(nh)/(2pi)=(3h)/(2pi) :. N=3`
`r_(n)=(in_(0)n^(2)h^(2))/(pimZe^(2))=(n^(2))/Z((in_(0)h^(2))/(Time^(2)))=(n^(2))/Z a_(0)`
`4.5a_(0)=(n^(2))/Z a_(0)`
`(because (in_(0)h^(2))/(pime^(2))=a_(0) (Bohr's radius))`
When Z=2
`1/lambda=RZ^(2)[1/(n_(f)^(2))-1/(n_(i)^(2))]=4R[1/(n_(f)^(2))-1/(n_(i)^(2))]`
When atom deexcites form 3rd to 1st shell
`1/(lambda_(1))=4R[1/1-1/9]=(32R)/9 :. lambda_(1)=9/32R`
When atom deexcites form 3rd to 2nd shell
`1/(lambda_(2))=4R[1/4-1/9]=(5R)/9 :. lambda_(2)=9/5R`