Question:
The stopping potential \( V_0 \) measured in a photoelectric experiment for a metal surface is plotted against frequency \( \nu \) of the incident radiation. Let \( m \) be the slope of the straight line so obtained. Then the value of charge of an electron is given by (\( h \) is the Planck's constant.)
The stopping potential \( V_0 \) measured in a photoelectric experiment for a metal surface is plotted against frequency \( \nu \) of the incident radiation. Let \( m \) be the slope of the straight line so obtained. Then the value of charge of an electron is given by (\( h \) is the Planck's constant.)
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Write \( eV_0 = h\nu - \phi \) and compare it with a straight line. The slope is \( h/e \).
Updated On: Oct 27, 2026
- \( mh \)
- \( \frac{m}{h} \)
- \( \dfrac{h}{m} \)
- \( \dfrac{1}{mh} \)
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The Correct Option is C
Solution and Explanation
Step 1: Write the photoelectric equation:
Einstein's equation says the maximum kinetic energy of an emitted electron is \( K_{max} = h\nu - \phi \), where \( \phi \) is the work function of the metal.
The stopping potential \( V_0 \) is the potential that just stops the fastest electron, so \( eV_0 = K_{max} \).
Step 2: Get \( V_0 \) as a function of \( \nu \):
Put \( K_{max} = eV_0 \) in the equation.
\[ eV_0 = h\nu - \phi \]
Divide by \( e \).
\[ V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi}{e} \]
Step 3: Compare with a straight line:
This is of the form \( y = mx + c \) with \( y = V_0 \) and \( x = \nu \).
So the slope of the \( V_0 \) versus \( \nu \) graph is \( m = \dfrac{h}{e} \).
Step 4: Solve for the charge:
From \( m = h/e \) we get
\[ e = \frac{h}{m} \]
This is option (C).
Step 5: Why the other options are wrong:
Option (A) \( mh \) equals \( h^2/e \), not \( e \).
Option (B) \( m/h \) equals \( 1/e \), which is the reciprocal of the charge.
Option (D) \( 1/(mh) \) equals \( e/h^2 \), which is also not \( e \).
Final Answer:
The charge of the electron is \( e = h/m \), option (C).
\[ \boxed{e = \dfrac{h}{m}} \]
Einstein's equation says the maximum kinetic energy of an emitted electron is \( K_{max} = h\nu - \phi \), where \( \phi \) is the work function of the metal.
The stopping potential \( V_0 \) is the potential that just stops the fastest electron, so \( eV_0 = K_{max} \).
Step 2: Get \( V_0 \) as a function of \( \nu \):
Put \( K_{max} = eV_0 \) in the equation.
\[ eV_0 = h\nu - \phi \]
Divide by \( e \).
\[ V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi}{e} \]
Step 3: Compare with a straight line:
This is of the form \( y = mx + c \) with \( y = V_0 \) and \( x = \nu \).
So the slope of the \( V_0 \) versus \( \nu \) graph is \( m = \dfrac{h}{e} \).
Step 4: Solve for the charge:
From \( m = h/e \) we get
\[ e = \frac{h}{m} \]
This is option (C).
Step 5: Why the other options are wrong:
Option (A) \( mh \) equals \( h^2/e \), not \( e \).
Option (B) \( m/h \) equals \( 1/e \), which is the reciprocal of the charge.
Option (D) \( 1/(mh) \) equals \( e/h^2 \), which is also not \( e \).
Final Answer:
The charge of the electron is \( e = h/m \), option (C).
\[ \boxed{e = \dfrac{h}{m}} \]
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