Slope of `V_(0)` vs v curve is (where `V_(0)`= Stopping potential, v=subjected freqency)
Slope of `V_(0)` vs v curve is (where `V_(0)`= Stopping potential, v=subjected freqency)
A
`e`
B
`h/e`
C
`phi`
D
`h`
Text Solution
AI Generated Solution
The correct Answer is:
To find the slope of the \( V_0 \) versus \( \nu \) curve, where \( V_0 \) is the stopping potential and \( \nu \) is the subjected frequency, we can follow these steps:
### Step-by-Step Solution:
1. **Understand the Photoelectric Effect**:
When light of frequency \( \nu \) strikes a metal surface, it can eject electrons if the energy of the light (given by \( h\nu \), where \( h \) is Planck's constant) is greater than the work function \( \phi \) (denoted as \( h\nu_0 \)) of the metal.
2. **Energy Equation**:
The energy of the incident photons can be expressed as:
\[
h\nu = \phi + KE
\]
where \( KE \) is the kinetic energy of the emitted electrons.
3. **Relate Kinetic Energy to Stopping Potential**:
The kinetic energy of the emitted electrons can also be expressed in terms of the stopping potential \( V_0 \):
\[
KE = eV_0
\]
where \( e \) is the charge of the electron.
4. **Combine the Equations**:
By substituting \( KE \) into the energy equation, we get:
\[
h\nu = h\nu_0 + eV_0
\]
5. **Rearranging the Equation**:
Rearranging the equation to express \( V_0 \) in terms of \( \nu \):
\[
eV_0 = h\nu - h\nu_0
\]
Dividing through by \( e \):
\[
V_0 = \frac{h}{e} \nu - \frac{h\nu_0}{e}
\]
6. **Identify the Linear Relationship**:
The equation \( V_0 = \frac{h}{e} \nu - \frac{h\nu_0}{e} \) is in the form of \( y = mx + c \), where:
- \( y = V_0 \)
- \( x = \nu \)
- \( m = \frac{h}{e} \) (the slope)
- \( c = -\frac{h\nu_0}{e} \) (the y-intercept)
7. **Conclusion**:
Therefore, the slope of the \( V_0 \) versus \( \nu \) curve is:
\[
\text{slope} = \frac{h}{e}
\]
### Final Answer:
The slope of the \( V_0 \) versus \( \nu \) curve is \( \frac{h}{e} \).
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