Question

(a) What is the wavelength of a photon that has a momentum of\({\bf{5}}{\bf{.00 \times 1}}{{\bf{0}}^{{\bf{ - 29}}}}{\bf{ kg}} \cdot {\bf{m/s}}\)? (b) Find its energy in eV.

Short answer

1. The wavelength is obtained as:\({\rm{1}}{\rm{.32 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ m}}\).
2. The energy in eV is obtained as: \({\rm{9}}{\rm{.39 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ eV}}\).

Step-by-step solution

Define Quantum Physics

The study of matter and energy at the most fundamental level is known as quantum physics. Its goal is to learn more about the characteristics and behaviors of nature's fundamental building elements.

Evaluating the wavelength

(a)

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

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

Here, the value of\(h\) is the Plank's constant and \(\rho \) is the momentum of the photon.

\(\begin{aligned}{}\lambda &= \frac{{6.62 \times {{10}^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}}}{{5.00 \times {{10}^{ - 29}}\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}}\\ &= 1.32 \times {10^{ - 5}}\;{\rm{m}}\end{aligned}\)

Therefore, the wavelength is\(1.32 \times {10^{ - 5}}{\rm{ m}}\).

Evaluating the energy in eV

(b)

The energy of a photon is written in the terms of its momentum in the form of equation:\(E = pc\).

Here the value of \(c\) is the speed of light. Then the conversion of energy from\({\rm{J}}\) to \({\rm{MeV}}\), one iss supposed to multiply the value by:

\({\rm{6}}{\rm{.24}} \times {\rm{1}}{{\rm{0}}^{{\rm{18}}}}\;\frac{{{\rm{eV}}}}{{\rm{J}}}\).

\(\begin{aligned}{}E{\rm{ }} &= {\rm{ }}6.62 \times {10^{ - 29}}\;\frac{{{\rm{kg}} \cdot {\rm{m}}}}{{\rm{s}}} \times 3.00 \times {10^8}\;{\rm{m}}{{\rm{s}}^{ - 1}}\\ &= {\rm{ }}1.50\times {10^{ - 20}}\;{\rm{J}}\\ &= {\rm{ }}1.50 \times {10^{ - 20}}\;{\rm{J}} \times 6.24 \times {10^{18}}\;\frac{{{\rm{eV}}}}{{\rm{J}}}\\ &= {\rm{ }}9.39 \times {10^{ - 2}}\;{\rm{eV}}\end{aligned}\)

Therefore, the energy in eV is: \({\rm{9}}{\rm{.39 \times 1}}{{\rm{0}}^{ - {\rm{2}}}}\;{\rm{eV}}\).