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Question

The radius of an orbit in a Hydrogen - like atom is 4.5 a0, where a0 is the Bohr radius. Its orbital angular momentum is 3h2π, it is given that h is Planck's constant and R is Rydberg constant. The possible wavelength when the atom de-excites from the given state to the ground state is:

A
932R
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B
916R
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C
95R
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D
43R
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Solution

The correct option is A 932R
Angular momentum =nh2π=3h2π
n=3
Also r=n2Za0
n2Za0=9a02=32a02
Z=2
Now de-excitation is occuring from n=3
to Ground state; i.e n=1
For de-excitation, 1λ=RZ2[1n211n22]
1λ=R×Z2×(112132)
1λ=4R(919)
1λ=4R×89
1λ=32R9
λ=932R

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