Chapter 3, Problem 1Q

To determine

The energy of photons that would be chosen and the reason.

Answer to Problem 1Q

The energy of the light must be greater than $1.2408\times {10}^8\text{eV}$ as this energy corresponds to the wavelength of light less than the size of the nucleus.

Explanation of Solution

The diffraction of photons by the nuclei occurs when the wavelength of the light is either equal to or less than the size of the nuclei.

Write the expression for the energy of the photon.

    $E>\frac{hc}{\lambda }$        (1)

Here, $\lambda$ is the wavelength of light, $h$ is the plank’s constant and $c$ is the speed of light.

Substitute $4.136\times {10}^{-15}\text{eV}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $10\times {10}^{-15}\text{m}$ for $\lambda$ in equation (1).

    $\begin{array}{c}E>\frac{\left(4.136\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(10\times {10}^{-15}\text{m}\right)}\\ >1.2408\times {10}^8\text{eV}\end{array}$

From above, it is clear that the energy of the photons must be greater than $1.2408\times {10}^8\text{eV}$.

Conclusion:

Therefore, the energy of the light must be greater than $1.2408\times {10}^8\text{eV}$ as this energy corresponds to the wavelength of light less than the size of the nucleus.