Representing the stopping potential V along y-axis and `((1)/(lamda))` along x-axis for a given photocathode, the curve is a straight line, the slope of which is equal to

Options

A. `(e)/(hc)`
B. `(hc)/(e)`
C. `(ec)/(h)`
D. `(e)/(W)`

Explanation

To solve the problem step by step, we need to analyze the relationship between the stopping potential \( V \) and the wavelength \( \lambda \) of the incident light in the context of the photoelectric effect.

### Step 1: Understand the photoelectric effect equation
The maximum kinetic energy \( K.E. \) of the emitted photoelectrons is given by:
\[ K.E. = h\nu - W \]
where:
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( W \) is the work function of the photocathode.

### Step 2: Relate frequency to wavelength
The frequency \( \nu \) can be expressed in terms of the wavelength \( \lambda \) using the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light.

### Step 3: Substitute frequency in the kinetic energy equation
Substituting \( \nu \) in the kinetic energy equation gives:
\[ K.E. = h \left(\frac{c}{\lambda}\right) - W \]

### Step 4: Relate kinetic energy to stopping potential
The kinetic energy of the emitted electrons is also related to the stopping potential \( V \) by:
\[ K.E. = eV \]
where \( e \) is the charge of the electron.

### Step 5: Set the equations equal
Equating the two expressions for kinetic energy, we have:
\[ eV = h\left(\frac{c}{\lambda}\right) - W \]

### Step 6: Rearrange the equation
Rearranging the equation to express \( V \) in terms of \( \frac{1}{\lambda} \):
\[ eV = \frac{hc}{\lambda} - W \]
\[ V = \frac{hc}{e} \cdot \frac{1}{\lambda} - \frac{W}{e} \]

### Step 7: Identify the slope of the line
From the equation \( V = \frac{hc}{e} \cdot \frac{1}{\lambda} - \frac{W}{e} \), we can see that this is in the form of \( y = mx + b \), where:
- \( y \) is \( V \),
- \( x \) is \( \frac{1}{\lambda} \),
- \( m \) (the slope) is \( \frac{hc}{e} \),
- \( b \) (the y-intercept) is \( -\frac{W}{e} \).

### Conclusion
Thus, the slope of the line representing the stopping potential \( V \) versus \( \frac{1}{\lambda} \) is:
\[ \text{slope} = \frac{hc}{e} \]