Question

(a) Find the momentum of a 100 - keV x-ray photon.

(b) Find the equivalent velocity of a neutron with the same momentum.

(c) What is the neutron's kinetic energy in keV?

Short answer

(a) The momentum of a photon is $$5.33\times 10^{-23}\mathrm{K}\mathrm{g}\times \mathrm{m}\times {\mathrm{s}}^{-1}$$.

(b) The velocity of neutrons is $$3.19\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$$.

(c) The kinetic energy of neutron is $$5.3\times 10^{-3}\mathrm{k}\mathrm{e}\mathrm{V}$$.

Step-by-step solution

Definition of Concept

Wavelength: The distance between identical points (adjacent crests) in adjacent cycles of a waveform signal propagated in space or along a wire is defined as the wavelength. This length is typically specified in wireless systems in metres (m), centimetres (cm), or millimetres (mm) (mm).

Find the momentum of a photon

(a)

Considering the given information,

$$\begin{array}{l}E\approx 100\mathrm{k}\mathrm{e}\mathrm{V}\\ c\approx 3\times {10}^8\mathrm{m}{\mathrm{s}}^{-1}\end{array}$$

Apply the formula,

The relationship between energy and velocity in proton is :

$$E\approx pc$$

Or

$$p\approx \frac{E}{c}$$

We get by substituting the given joule values in the eV relation.

$$1\mathrm{e}\mathrm{V}=1.6\times 10^{-19}\mathrm{J}$$

Currently, the photon's energy is,

$$\begin{array}{l}E\approx 100\times {10}^3\times 1.6\times {10}^{-19}\mathrm{e}\mathrm{V}\\ E\approx 1.6\times {10}^{-15}\mathrm{J}\end{array}$$

We obtain the photon's momentum,

$$\begin{array}{l}p\approx \frac{1.6\times {10}^{-14}}{3\times {10}^8}\\ p\approx 5.33\times {10}^{-23}\mathrm{k}\mathrm{g}\cdot \mathrm{m}\cdot {\mathrm{s}}^{-1}\end{array}$$

Therefore, the required momentum of a photon is $$5.33\times 10^{-23}\mathrm{K}\mathrm{g}\times \mathrm{m}\times {\mathrm{s}}^{-1}$$.

Find the velocity of neutrons

(b)

Considering the given information,

$$\begin{array}{l}m\approx 1.67\times {10}^{-27}\mathrm{k}\mathrm{g}\\ \mathrm{p}=5.33\times 10^{-23}\mathrm{k}\mathrm{g}\times \mathrm{m}\times {\mathrm{s}}^{-1}\end{array}$$

Apply the formula,

The relationship between photon and wavelength for photon momentum is:

$$p\approx mv$$

Or

$$v\approx \frac{p}{m}$$

Currently, the photon's velocity is:

$$\begin{array}{l}v\approx \frac{5.33\times {10}^{-23}}{1.67\times {10}^{-27}}\\ v\approx 3.19\times {10}^4\mathrm{m}{\mathrm{s}}^{-1}\end{array}$$

Therefore, the velocity of neutrons is $$3.19\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$$.

Find the kinetic energy of neutron

(c)

Considering the given information,

$$\begin{array}{l}m\approx 1.67\times {10}^{-27}\mathrm{k}\mathrm{g}\\ v\approx 3.19\times {10}^4\mathrm{m}{\mathrm{s}}^{-1}\end{array}$$

Apply the formula,

The relationship between energy and velocity in proton kinetic energy is

$$KE\approx \frac{1}{2}m{v}^2$$

We get by putting the value for electron kinetic energy.

$$\begin{array}{l}KE\approx \frac{1}{2}\times 1.67\times {10}^{-27}\times ({}_{\times }^{3.19}\mathrm{J}\\ KE\approx 8.52\times {10}^{-19}\mathrm{J}\end{array}$$

We get by substituting the given joule values in the eV relation.

$$1\mathrm{e}\mathrm{V}=1.6\times 10^{-19}\mathrm{J}$$

Kinetic energy is measured in joules and is written as,

$$\begin{array}{l}E\approx 8.52\times {10}^{-19}\times \frac{1}{1.6\times {10}^{-19}}\mathrm{e}\mathrm{V}\\ E\approx 5.3\times {10}^{-3}\mathrm{k}\mathrm{e}\mathrm{V}\end{array}$$

Therefore, the requiredkinetic energy of neutron is$$5.3\times 10^{-3}\mathrm{k}\mathrm{e}\mathrm{V}$$.