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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)A. `(9)/(32R)`B. `(9)/(16R)`C. `(9)/(5R)`D. `(4)/(3R)` |
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Answer» Correct Answer - A::C Given data `4.5_(alpha_0)=alpha_0(n^2)/(Z)` `(nh)/(2pi)=(3h)/(2pi)` So `n=3` and `z=2` So possible wavelength are `(1)/(lamda_1)=RZ^2[(1)/(1^2)-(1)/(3^2)]implieslamda_1=(9)/(32R)` `(1)/(lamda_2)=RZ^2[(1)/(1^2)-(1)/(2^2)]implieslamda_2=(1)/(3R)` `(1)/(lamda_3)=RZ^2[(1)/(2^2)-(1)/(3^2)]implieslamda_3=(9)/(5R)` |
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