Relation between Kinetic Energy and Wavelength:

 
K.E (E) =1/2 mv2
v =  √2E / m ⇒ λ = h/ mv         
λ = h / √2E.m

 

Illustration:

An e-, a proton and an alpha particle have K.E of 16 E, 4 E and E respectively. What’s the qualitative order of their Broglie wavelengths?

(A)    λe > λP > λa                (B)    λe > λp = λa

(C)    λP < λe < λa                (D)    λa < λe = λP

Solution:

λ = h / √2mK.E . Hence (B) is correct.
 
Illustration:

What is the de-Broglie wavelength of electron having K.E. of 5 eV?

Solution:

K.E. = 1/2 mv2

 

 

v =√ 2K.E / m

Now = λ = h / mv 

= h / √2K.E.m

= 6.62 x 10-34 / √ 2 x 5 x 1.6 x 10-19 x 9.1 x 10-31

= 5.486 x 10-10 m
 
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