(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?

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).

(a)

Considering the given information,

\[\begin{array}{l}E \approx 100{\rm{keV}}\\c \approx 3 \times {10^8}\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\]

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.

\[{\rm{1eV = 1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{\;J}}\]

Currently, the photon's energy is,

\[\begin{array}{l}E \approx 100 \times {10^3} \times 1.6 \times {10^{ - 19}}{\rm{eV}}\\E \approx 1.6 \times {10^{ - 15}}\;{\rm{J}}\end{array}\]

We obtain the photon's momentum,

\[\begin{array}{l}p \approx \frac{{1.6 \times {{10}^{ - 14}}}}{{3 \times {{10}^8}}}\\p \approx 5.33 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\end{array}\]

Therefore, the required momentum of a photon is \[{\rm{5}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}{\rm{Kg \times m \times }}{{\rm{s}}^{{\rm{ - 1}}}}\].

(b)

Considering the given information,

\[\begin{array}{l}m \approx 1.67 \times {10^{ - 27}}\;{\rm{kg}}\\{\rm{p = 5}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}{\rm{\;kg \times m \times }}{{\rm{s}}^{{\rm{ - 1}}}}\end{array}\]

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:

\[\begin{array}{l}v \approx \frac{{5.33 \times {{10}^{ - 23}}}}{{1.67 \times {{10}^{ - 27}}}}\\v \approx 3.19 \times {10^4}\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\]

Therefore, the velocity of neutrons is \[{\rm{3}}{\rm{.19 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{\;m}}{{\rm{s}}^{{\rm{ - 1}}}}\].

(c)

Considering the given information,

\[\begin{array}{l}m \approx 1.67 \times {10^{ - 27}}\;{\rm{kg}}\\v \approx 3.19 \times {10^4}\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\]

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.

\[\begin{array}{l}KE \approx \frac{1}{2} \times 1.67 \times {10^{ - 27}} \times \left( {_ \times ^{3.19}\;{\rm{J}}} \right.\\KE \approx 8.52 \times {10^{ - 19}}\;{\rm{J}}\end{array}\]

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

\[{\rm{1eV = 1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}{\rm{\;J}}\]

Kinetic energy is measured in joules and is written as,

\[\begin{array}{l}E \approx 8.52 \times {10^{ - 19}} \times \frac{1}{{1.6 \times {{10}^{ - 19}}}}{\rm{eV}}\\E \approx 5.3 \times {10^{ - 3}}{\rm{keV}}\end{array}\]

Therefore, the requiredkinetic energy of neutron is\[{\rm{5}}{\rm{.3 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{keV}}\].