Question

(a) Calculate the momentum of a photon having a wavelength of \({\bf{2}}{\bf{.50}}\;{\bf{\mu m}}\).

(b) Find the velocity of an electron having the same momentum.

(c) What is the kinetic energy of the electron, and how does it compare with that of the photon?

Short answer

a. The momentum of photon is \({\rm{2}}{\rm{.65 \times 1}}{{\rm{0}}^{ - {\rm{28}}}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\).

b. The velocity of an electron having the same momentum is \({\rm{291\;m}}{{\rm{s}}^{{\rm{ - 1}}}}\).

c. The kinetic energy of an electron is\({\rm{3}}{\rm{.86 \times 1}}{{\rm{0}}^{ - {\rm{26}}}}{\rm{\;J}}\),energy of photon is \({\rm{7}}{\rm{.96}} \times {\rm{1}}{{\rm{0}}^{ - {\rm{20}}}}\) J and ratio their ratio is \({\rm{2}}{\rm{.06 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{\;J}}\).

Step-by-step solution

Determine the formulas:

To evaluate the wavelength, of the value\({\rm{\lambda }}\), of a photon with a given momentum. Consider the equation of the wavelength:

\({\bf{\lambda = }}\frac{{\bf{h}}}{{\bf{\rho }}}\)

Consider the formula for the momentum is:

\({\bf{p}} = {\bf{mv}}\)

Consider the formula for the kinetic energy:

\({\bf{KE}} = \frac{{\bf{1}}}{{\bf{2}}}{\bf{m}}{{\bf{v}}^{\bf{2}}}\)

Find the momentum of photon

(a)

Considering the given information,

\(\begin{array}{l}\lambda = 2.50\;{\rm{\mu m}}\\h = 6.63 \times {10^{ - 34}}\;{\rm{J}}{{\rm{s}}^{ - {\rm{1}}}}\end{array}\)

Apply the formula,

The relationship between photon and wavelength for photon momentum is:

\(p \approx \frac{h}{\lambda }\)

If the wavelength value is in micrometers, convert it to\({\rm{m}}\):

\(\lambda = 2.50 \times {10^{ - 6}}\;{\rm{m}}\)

Currently, the photon's momentum is,

\(\begin{array}{l}p \approx \frac{{6.63 \times {{10}^{ - 34}}}}{{2.50 \times {{10}^{ - 6}}}}\\p \approx 2.65 \times {10^{ - 28}}\;\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{{\rm{ - 1}}}}\end{array}\)

Therefore, the required momentum of photon is \({\rm{2}}{\rm{.65 \times 1}}{{\rm{0}}^{ - {\rm{28}}}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\).

Find the velocity of an electron

(b)

Considering the given information,

\(\begin{array}{l}p = 2.65 \times {\rm{1}}{{\rm{0}}^{ - {\rm{28}}}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\\m = 9.11 \times {10^{ - 31}}\;\;{\rm{kg}}\end{array}\)

Apply the formula,

The relationship between photon and velocity for momentum in photons is

Consider the classical expression to find the velocity of an electron with this momentum because it is indeed small.

\(p \approx mv\)

\(v \approx \frac{p}{m}\)

Currently, the photon's velocity is calculate as:

\(\begin{array}{l}v \approx \frac{{2.65 \times {{10}^{ - 28}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}}}{{9.11 \times {{10}^{ - 31}}\;\;{\rm{kg}}}}\\v \approx 291\;\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\)

Therefore, the velocity of an electron having the same momentum is \({\rm{291\;m}}{{\rm{s}}^{ - {\rm{1}}}}\).

Find the kinetic energy of an electron

(c)

Considering the given information,

\(\begin{array}{l}p = 2.65 \times {\rm{1}}{{\rm{0}}^{ - {\rm{28}}}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\\m = 9.11 \times {10^{ - 31}}\;\;{\rm{kg}}\end{array}\)

Apply the formula,

The relationship between energy and velocity in photons is:

\(E \approx pc\)

The relationship between energy and wavelength in electron kinetic energy is,

\(KE \approx \frac{1}{2}m{v^2}\)

Substitute the value for electron kinetic energy.

\(\begin{array}{l}KE \approx \frac{1}{2} \times 9.11 \times {10^{ - 31}} \times {()^{291}}\;{\rm{J}}\\KE \approx 3.86 \times {10^{ - 26}}\;{\rm{J}}\end{array}\)

Currently, photon's energy is:

\(\begin{array}{l}E \approx 2.65 \times {10^{ - 28}} \times 3 \times {10^8}\;{\rm{J}}\\E \approx 7.95 \times {10^{ - 20}}\;{\rm{J}}\end{array}\)

The ratio of electron kinetic energy to photon energy is:

\(\begin{array}{c}{\rm{ Ratio }} \approx \frac{{7.95 \times {{10}^{ - 20}}}}{{3.86 \times {{10}^{ - 26}}}}\\{\rm{Ratio }} \approx 2.06 \times {10^6}\end{array}\)

Therefore, the kinetic energy of an electron is\({\rm{3}}{\rm{.86 \times 1}}{{\rm{0}}^{{\rm{ - 26}}}}{\rm{\;J}}\),energy of photon is \({\rm{7}}{\rm{.96 \times 1}}{{\rm{0}}^{{\rm{ - 20}}}}\) J and ratio their ratio is \({\rm{2}}{\rm{.06 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{\;J}}\).