Question

# An excited electron of $$H$$-atoms emits a photon of wavelength $$\lambda$$ and returns in the ground state, the principal quantum number of excited state is given by:

A
λR(λR1)
B
λR(λR1)
C
1λR(λR1)
D
(λR1)λR

Solution

## The correct option is D $$\sqrt { \dfrac { \lambda R }{ \left( \lambda R-1 \right) } }$$we know that,  $$\cfrac{1}{\lambda}=RZ^2\left(\cfrac{1}{n_L^2}-\cfrac{1}{n_H^2}\right)$$The hydrogen atom returns to the ground state$${ n }_{ L }^{ 2 }=1$$ and Z for hydrogen atom=1Substituting in the equation, we get$$\cfrac{1}{\lambda}=RZ^2\left(\cfrac{1}{1^2}-\cfrac{1}{n_H^2}\right)$$$$\cfrac{1}{\lambda}=R.1.\left(1-\cfrac{1}{n^2}\right)$$$$\cfrac{1}{\lambda}=R\left(1-\cfrac{1}{n^2}\right)$$$$\cfrac { 1 }{ \lambda R } =R\left( 1-\cfrac { 1 }{ n^{ 2 } } \right)$$$$\cfrac { 1 }{ \lambda R } = 1-\cfrac { 1 }{ n^{ 2 } }$$$$\cfrac { 1 }{ { n }^{ 2 } } =1-\cfrac { 1 }{ \lambda R }$$$$\cfrac { 1 }{ { n }^{ 2 } } =\cfrac { \lambda R-1 }{ \lambda R }$$$$n=\sqrt { \cfrac { \lambda R }{ \lambda R-1 } }$$So, the correct option is $$B$$Chemistry

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