Question

(a) What is the momentum of a$\mathbf{0}\mathbf{.0100}\mathbf{-}\mathbf{n}\mathbf{m}$-wavelength photon that could detect details of an atom? (b) What is its energy in MeV?

Short answer

1. The momentum is obtained as $6.62\times {10}^{-23}\mathrm{k}\mathrm{g}\cdot \mathrm{m}\cdot {\mathrm{s}}^{-1}$.
2. The energy in MeV is obtained as $0.124\mathrm{M}\mathrm{e}\mathrm{V}$.

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.

Evaluate the momentum

(a)

The momentum of the value $\rho$ of the photon is written in terms of the Plank's constant, of the value $h$, and the wavelength, of the value $\lambda$:

$\rho =\frac{h}{\lambda }$

Solve further as:

$\begin{array}{rl}\rho & =\frac{6.62\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}}{0.01\times {10}^{-9}\mathrm{m}}\\ & =6.62\times {10}^{-23}\mathrm{k}\mathrm{g}\cdot \mathrm{m}\cdot {\mathrm{s}}^{-1}\end{array}$

Therefore, the momentum is $6.62\times {10}^{-23}\mathrm{k}\mathrm{g}\cdot \mathrm{m}\cdot {\mathrm{s}}^{-1}$.

Evaluate the energy in MeV

(b)

Consider the expression for the energy of photon in terms of the momentum as:

$E=pc$

Here the value of $c$ is the speed of light. Then the conversion of energy from $J$to $\mathrm{M}\mathrm{e}\mathrm{V}$, we are suppose to multiply the value by: $6.24\times 10^{12}\frac{\mathrm{M}\mathrm{e}\mathrm{V}}{\mathrm{J}}$.

$\begin{array}{rl}E& =6.62\times {10}^{-23}\frac{\mathrm{k}\mathrm{g}\times \mathrm{m}}{\mathrm{s}}\times 3.00\times {10}^8\frac{\mathrm{m}}{\mathrm{s}}\\ & =1.99\times {10}^{-14}\mathrm{J}\\ & =1.99\times {10}^{-14}\mathrm{J}\times 6.24\times {10}^{12}\frac{\mathrm{M}\mathrm{e}\mathrm{V}}{\mathrm{J}}\\ & =0.124\mathrm{M}\mathrm{e}\mathrm{V}\end{array}$

Therefore, the energy in MeV is:$0.124\mathrm{M}\mathrm{e}\mathrm{V}$.