Question

If the electron revolves in n = 2 of H atom then how many de-broglie wavelengths are associated with it ?

Answer 1

Answer:

0.67nm

Explanation:

n=2

K.E energy of the second state is:

n  

2

 

13.6

​  

=  

4

13.6

​  

 

3.4×1.6×10  

−19

j

de brogile wavelength

λ=  

2mE  

k

​  

 

​  

 

h

​  

 

2×9.1×10  

−31

×3.4×1.6×10  

−19

 

​  

 

6.63×10  

−34

 

​  

 

=0.67nm