Relation between Kinetic Energy and Wavelength:
K.E (E) =1/2 mv2
v = √2E / m ⇒ λ = h/ mv
λ = h / √2E.m
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
λ = h / √2mK.E . Hence (B) is correct.
What is the de-Broglie wavelength of electron having K.E. of 5 eV?
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