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) λ_e > λ_P > λ_a (B) λ_e > λ_p = λ_a

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