Chapter 44, Problem 1P

(a)

To determine

The magnitude of minimum frequency.

Answer to Problem 1P

The minimum frequency is $2.27\times {10}^{23}\text{Hz}$.

Explanation of Solution

In this case, each photon would have same energy that is the rest mass energy of the proton to conserve momentum. The kinetic energy can be neglected in this case.

Write the expression to calculate the minimum frequency.

  $f=\frac{E_0}{h}$

Here, f is the minimum frequency, $E_0$ is the rest mass energy of proton and h is the Planck’s constant.

Substitute $938.3\text{MeV}$ for $E_0$ and $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for h in the above equation to calculate f.

  $\begin{array}{c}f=\frac{938.3\text{MeV}\left(\frac{{10}^6\text{eV}}{1\text{MeV}}\right)\left(\frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{6.626\times {10}^{-34}\text{J}\cdot \text{s}}\\ =2.27\times {10}^{23}\text{Hz}\end{array}$

Conclusion:

Therefore, the minimum frequency is $2.27\times {10}^{23}\text{Hz}$.

(b)

To determine

The wavelength of each photon.

Answer to Problem 1P

The wavelength of each photon is $1.32\times {10}^{-15}\text{m}$.

Explanation of Solution

Write the expression to calculate the wavelength.

  $\lambda =\frac{c}{f}$

Here, $\lambda$ is the wavelength,

Substitute $3.00\times {10}^8\text{m}/\text{s}$ for c and $2.27\times {10}^{23}\text{Hz}$ for f in the above equation to calculate $\lambda$.

  $\begin{array}{c}\lambda =\frac{3.00\times {10}^8\text{m}/\text{s}}{2.27\times {10}^{23}\text{Hz}}\\ =1.32\times {10}^{-15}\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of each photon is $1.32\times {10}^{-15}\text{m}$.