A photon and an electron have equal energy `E . lambda_("photon")//lambda_("electron")` is proportional to

Options

A. `sqrt(E)`
B. `1/sqrt(E)`
C. `1/E`
D. does not depend upon E

Answer

To solve the problem, we need to determine the relationship between the wavelengths of a photon and an electron when both have equal energy \( E \).

### Step 1: Energy of the Photon
The energy of a photon is given by the equation:
\[ E = h \nu \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the photon. The wavelength \( \lambda \) of the photon is related to its frequency by the equation:
\[ \lambda = \frac{c}{\nu} \]
where \( c \) is the speed of light. Therefore, we can express the energy of the photon in terms of its wavelength:
\[ E = \frac{hc}{\lambda_{\text{photon}}} \]

### Step 2: Energy of the Electron
For an electron, we can use the non-relativistic kinetic energy formula:
\[ E = \frac{mv^2}{2} \]
where \( m \) is the mass of the electron and \( v \) is its velocity. The de Broglie wavelength \( \lambda \) of the electron is given by:
\[ \lambda_{\text{electron}} = \frac{h}{mv} \]

### Step 3: Relating Energy to Wavelength for the Electron
From the kinetic energy equation, we can express the velocity \( v \) in terms of energy \( E \):
\[ v = \sqrt{\frac{2E}{m}} \]
Substituting this expression for \( v \) into the de Broglie wavelength equation gives:
\[ \lambda_{\text{electron}} = \frac{h}{m \sqrt{\frac{2E}{m}}} = \frac{h}{\sqrt{2Em}} \]

### Step 4: Finding the Ratio of Wavelengths
Now we have expressions for both wavelengths:
1. For the photon:
\[ \lambda_{\text{photon}} = \frac{hc}{E} \]
2. For the electron:
\[ \lambda_{\text{electron}} = \frac{h}{\sqrt{2Em}} \]

Now we can find the ratio:
\[ \frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{\frac{hc}{E}}{\frac{h}{\sqrt{2Em}}} = \frac{hc \sqrt{2Em}}{E} \]

### Step 5: Simplifying the Ratio
This can be simplified to:
\[ \frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc \sqrt{2m}}{E^{1/2}} \]
This shows that the ratio of the wavelengths is proportional to \( \sqrt{E} \).

### Conclusion
Thus, we conclude that:
\[ \lambda_{\text{photon}} \propto \sqrt{E} \]
and
\[ \lambda_{\text{electron}} \propto \sqrt{E} \]

Therefore, the answer is that \( \lambda_{\text{photon}} / \lambda_{\text{electron}} \) is proportional to \( \sqrt{E} \).

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