Question

Two gases He and Ne at 127 C and 27 C , calculate ratio of their de broglie's wavelength for Ne : He

Answer 1

Answer:

Explanation:

de-Broglie wavelength λ= $\frac{h}{mv}$

Where the speed (r.m.s) of a gas particle at the given temperature (T) is given

as

$\frac{1}{2}mv^{2} = \frac{3}{2} kT$

⇒ $v = \sqrt{} \frac{3KT}{m}$

where k= Boltzmanns constant and m= mass of the gas particle and T= temperature of the gas in k

⇒$mv = \sqrt{3mKT}$

⇒ λ $= \frac{h}{mv} = \frac{x}{\sqrt{3mKt} }$

⇒mv=  

3mKT

​

 

⇒λ=  

mv

h

​

=  

3mkT

​

 

h

​

 

∴  

λ  

He

​

 

λ  

H

​

 

​

=  

m  

H

​

 T  

H

​

 

m  

He

​

 T  

He

​

 

​

 

​

 

=  

(2 amu) (273+127)  

∘

k

(4 amu) (273+127)  

∘

k

​

 

​

=  

3

8

​

 

​