In the intercept on the y-axis is equal to

Options

A. `+(W)/(e)`
B. `-(W)/(e)`
C. `-We`
D. `(e)/(W)`

Explanation

To solve the question about the intercept on the y-axis in the context of the photoelectric effect, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Maximum Kinetic Energy of Photoelectrons**:
The maximum kinetic energy (K.E.) of photoelectrons emitted during the photoelectric effect can be expressed as:
\[ K.E. = \frac{1}{2} mv^2 = h\nu - W \]
where \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(W\) is the work function of the material.

2. **Relating Frequency to Wavelength**:
The frequency \(\nu\) can be related to the wavelength \(\lambda\) of the light by the equation:
\[ \nu = \frac{c}{\lambda} \]
where \(c\) is the speed of light.

3. **Substituting Frequency in the K.E. Equation**:
Substituting the expression for \(\nu\) into the K.E. equation gives:
\[ K.E. = h\left(\frac{c}{\lambda}\right) - W \]

4. **Expressing K.E. in Terms of Voltage**:
The kinetic energy can also be expressed in terms of electric potential \(V\) as:
\[ K.E. = eV \]
where \(e\) is the charge of the electron.

5. **Setting Up the Equation**:
Equating the two expressions for K.E. gives:
\[ eV = h\left(\frac{c}{\lambda}\right) - W \]

6. **Rearranging the Equation**:
Rearranging this equation to isolate \(V\) yields:
\[ V = \frac{hc}{e} \cdot \frac{1}{\lambda} - \frac{W}{e} \]

7. **Identifying the Intercept**:
The equation can be rewritten in the form of a straight line \(y = mx + b\), where:
- \(y\) corresponds to \(V\)
- \(x\) corresponds to \(\frac{1}{\lambda}\)
- The slope \(m\) is \(\frac{hc}{e}\)
- The intercept \(b\) on the y-axis is \(-\frac{W}{e}\)

8. **Conclusion**:
Therefore, the intercept on the y-axis is equal to:
\[ \text{Intercept} = -\frac{W}{e} \]

### Final Answer:
The intercept on the y-axis is equal to \(-\frac{W}{e}\).

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