Question

The energy of thermal neutrons is 0.02 eV, the
associated De broglie wave length is:-
(1) 2 A
(2) 0.2A
(3) 0.02 A
(4) 0.002A
​

Answer 1

Answer:

E = 1/2 m v2  

v = (2 E / m)^1/2

p = m v = (2 E m)^1/2   where p = momentum

wavelength = h / p = 6.62E-34 / (2 * .02 * 1.6E-19 * 1.67E-27)^1/2

since 1 eV = 1.6E-19 J  and m = 1.67E-27

wavelength = 2.02E-10 m = 2.02 A