Question

Repeat the previous problem for a \({\bf{10}}{\bf{.0}}\)-nm-wavelength photon.

Short answer

1. The momentum of photon is\(6.63 \times {10^{ - 26}}\;\;{\rm{kg}} \cdot {\rm{m}}{{\rm{s}}^{ - 1}}\).
2. The velocity of an electron having the same momentum is\({\rm{7}}{\rm{.27}} \times {\rm{1}}{{\rm{0}}^{\rm{4}}}{\rm{\;m}}{{\rm{s}}^{{\rm{ - 1}}}}\).
3. The kinetic energy of an electron is \({\rm{2}}{\rm{.40 \times 1}}{{\rm{0}}^{ - {\rm{21}}}}{\rm{\;J}}\), energy of photon is \(1.98 \times {10^{ - 17}}\) J and ratio their ratio is \({\rm{8}}{\rm{.25 \times 1}}{{\rm{0}}^{\rm{3}}}\).

Step-by-step solution

Define the formula of Concept

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,

\(\begin{array}{l}\lambda = 10.0\;{\rm{nm}}\\h = 6.63 \times {10^{ - 34}}\;{\rm{J}}{{\rm{s}}^{ - {\rm{1}}}}\end{array}\)

Apply the formula,

\(\lambda = 10 \times {10^{ - 9}}\;{\rm{m}}\)

Currently, the photon's momentum is,

\(\begin{array}{l}p \approx \frac{{6.63 \times {{10}^{ - 34}}}}{{10 \times {{10}^{ - 9}}}}\\p \approx 6.63 \times {10^{ - 26}}\;\;{\rm{kg}} \cdot {\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\)

Therefore, the required momentum of photon is \(6.63 \times {10^{ - 26}}\;\;{\rm{kg}} \cdot {\rm{m}}{{\rm{s}}^{ - 1}}\).

Find the velocity of an electron

(b)

Considering the given information,

\(\begin{array}{l}p = 6.63 \times {10^{ - 26}}\;\;{\rm{kg}} \cdot {\rm{m}}{{\rm{s}}^{ - 1}}\\m = 9.11 \times {10^{ - 31}}\;\;{\rm{kg}}\end{array}\)

Apply the formula,

The relationship between photon and velocity for momentum in photons is”:

\(p \approx mv\)

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

Currently, the photon's velocity is as follows:

\(\begin{array}{l}v \approx \frac{{6.63 \times {{10}^{ - 26}}\;{\rm{kg}} \cdot {\rm{m}}{{\rm{s}}^{ - 1}}}}{{9.11\;{\rm{kg}}}}\\v \approx 7.27 \times {10^4}\;\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\)

Therefore, the velocity of an electron having the same momentum is \({\rm{7}}{\rm{.27}} \times {\rm{1}}{{\rm{0}}^{\rm{4}}}\;{\rm{m}}{{\rm{s}}^{ - {\rm{1}}}}\).

Find the kinetic energy of an electron

(c)

Considering the given information:

\(\begin{array}{l}p = {\rm{6}}{\rm{.63 \times 1}}{{\rm{0}}^{ - {\rm{26}}}}{\rm{\;}}\;{\rm{kg}} \cdot {\rm{m}}{{\rm{s}}^{ - 1}}\\m = {\rm{9}}{\rm{.11 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}\;{\rm{kg}}\\v \approx 7.27 \times {10^4}\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\)

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.

\(\begin{array}{l}KE \approx \frac{1}{2} \times 9.11 \times {10^{ - 31}} \times \left( {7.27 \times {{10}^4}\;{\rm{m}}{{\rm{s}}^{ - 1}}} \right)\\KE \approx 2.40 \times {10^{ - 21}}\;{\rm{J}}\end{array}\)

Currently, photon's energy is:

\(\begin{array}{l}E \approx 6.65 \times {10^{ - 28}} \times 3 \times {10^8}\;{\rm{J}}\\E \approx 1.98 \times {10^{ - 17}}\;{\rm{J}}\end{array}\)

The ratio of electron kinetic energy to photon energy is:

\(\begin{array}{c}{\rm{ Ratio }} \approx \frac{{1.98 \times {{10}^{ - 17}}}}{{2.40 \times {{10}^{ - 21}}}}{\rm{ }}\\{\rm{Ratio }} \approx 8.25 \times {10^3}\end{array}\)

Therefore, the kinetic energy of an electron is\({\rm{2}}.40 \times {10^{ - 21}}{\rm{\;J}}\), energy of photon is \(1.98 \times {10^{ - 17}}\) J and ratio their ratio is \(8.25 \times {\rm{1}}{{\rm{0}}^{\rm{3}}}\).