Question

For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40 \times 10^{-10}$ m?

Answer 1

In this question, we will use the formula of kinetic energy which gives the relation with velocity. Then, we will use the relation of wavelength and momentum. After equating both the relations, we get kinetic energy in the form of mass and wavelength.

Complete step-by-step solution:
Given: The wavelength of the neutron, $\lambda = 1.40 \times 10^{-10}$ m.
We know, kinetic energy is given by:
$K = \dfrac{1}{2}m v^{2}$……($1$)
K is the kinetic energy.
m is the mass of a neutron.
v is the speed of a neutron.
Wavelength and momentum is related by:
$\lambda = \dfrac{h}{p}$
$\implies \lambda = \dfrac{h}{p}$ ……($2$)
Using equation ($1$) and ($2$),
$K = \dfrac{h^{2}}{2 \lambda^{2} m}$
Put $m = 1.66 \times 10^{-27}$ Kg, $\lambda = 1.40 \times 10^{-10}$ m and $h = 6.6 \times 10^{-34}$ Js in the above formula.
We get,
$K = \dfrac{(6.6 \times 10^{-34})^{2}}{2 (1.40 \times 10^{-10})^{2} (1.66 \times 10^{-27})}$
$\implies K = 6.75 \times 10^{-21}$ J
Now, we convert J to eV.
$K = \dfrac{6.75 \times 10^{-21}}{1.6 \times 10^{-19}} = 4.22 \times 10^{-2}$ eV.

Note: The amount of energy is directly proportional to the frequency and, thus, equivalently, is inversely proportional to the wavelength. The larger the frequency, the greater its energy. Equivalently, the higher the wavelength, the lower its energy.