The log - log graph between the energy `E` of an electron and its de - Broglie wavelength `lambda` will be

Options

A.
B.
C.
D.

Explanation

To analyze the relationship between the energy \( E \) of an electron and its de Broglie wavelength \( \lambda \), we can start from the de Broglie wavelength formula and derive the log-log relationship step by step.

### Step-by-Step Solution:

1. **Understanding the de Broglie Wavelength**:
The de Broglie wavelength \( \lambda \) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Relating Momentum to Energy**:
For an electron, the momentum \( p \) can be expressed in terms of its kinetic energy \( E \):
\[ p = \sqrt{2mE} \]
where \( m \) is the mass of the electron.

3. **Substituting Momentum into the Wavelength Formula**:
By substituting the expression for momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

4. **Taking the Logarithm of Both Sides**:
To analyze the relationship in a log-log graph, we take the logarithm of both sides:
\[ \log(\lambda) = \log\left(\frac{h}{\sqrt{2mE}}\right) \]

5. **Applying Logarithmic Properties**:
Using the properties of logarithms, we can separate the terms:
\[ \log(\lambda) = \log(h) - \log(\sqrt{2m}) - \log(\sqrt{E}) \]
This simplifies to:
\[ \log(\lambda) = \log(h) - \frac{1}{2}\log(2m) - \frac{1}{2}\log(E) \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ \log(\lambda) = \log(h) - \frac{1}{2}\log(2m) - \frac{1}{2}\log(E) \]
This can be expressed as:
\[ \log(\lambda) = -\frac{1}{2}\log(E) + \left(\log(h) - \frac{1}{2}\log(2m)\right) \]

7. **Identifying the Log-Log Graph**:
In the equation \( \log(\lambda) = -\frac{1}{2}\log(E) + C \), where \( C = \log(h) - \frac{1}{2}\log(2m) \), we can see that:
- The slope of the graph (in a log-log plot) is \( -\frac{1}{2} \).
- The intercept on the \( y \)-axis is \( C \).

### Conclusion:
The log-log graph between the energy \( E \) of an electron and its de Broglie wavelength \( \lambda \) will have a slope of \( -\frac{1}{2} \) and a positive intercept on the \( y \)-axis.