To solve this problem, let's first write down the relevant formulas and given information:
1. **Energy of a photon (\(E\))**:
The energy of a photon is given by:
\[
E = \frac{hc}{\lambda_2}
\]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda_2\) is the wavelength of the photon.
2. **Kinetic energy of the proton**:
We are given that the energy of the photon is equal to the kinetic energy of the proton. Therefore:
\[
K = E
\]
where \(K\) is the kinetic energy of the proton.
3. **de Broglie wavelength of the proton (\(\lambda_1\))**:
The de Broglie wavelength is given by:
\[
\lambda_1 = \frac{h}{p}
\]
where \(p\) is the momentum of the proton. The momentum \(p\) can be related to the kinetic energy \(K\) by:
\[
p = \sqrt{2mK}
\]
where \(m\) is the mass of the proton.
Now, let's proceed to find the relationship between \(\frac{\lambda_1}{\lambda_2}\).
### Step-by-step solution:
1. **Express the de Broglie wavelength of the proton (\(\lambda_1\))**:
Substitute \(p = \sqrt{2mK}\) into the de Broglie wavelength formula:
\[
\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mK}}
\]
2. **Express \(\lambda_1\) using \(E\)** (since \(K = E\)):
\[
\lambda_1 = \frac{h}{\sqrt{2mE}}
\]
3. **Express the wavelength of the photon (\(\lambda_2\))**:
From the energy of the photon:
\[
\lambda_2 = \frac{hc}{E}
\]
4. **Find the ratio \(\frac{\lambda_1}{\lambda_2}\)**:
\[
\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}
\]
Simplify the expression:
\[
\frac{\lambda_1}{\lambda_2} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \frac{E^{1/2}}{\sqrt{2m} \cdot c}
\]
From the above equation, it's clear that \(\frac{\lambda_1}{\lambda_2}\) is proportional to \(E^{1/2}\).
### Answer:
\[
\frac{\lambda_1}{\lambda_2} \propto E^{1/2}
\]
So, the correct option is (B) \(E^{1/2}\).
