CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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=1
Substituting 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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image