Question

In photoelectric effect, the stopping potential \( V_0 \) vs frequency \( \nu \) curve is plotted. \( h \) is the Planck's constant and \( \phi_0 \) is the work function of metal. 
(A) \( V_0 \) vs \( \nu \) is linear. 
(B) The slope of \( V_0 \) vs \( \nu \) curve is \( \frac{\phi_0}{h} \). 
(C) \( h \) constant is related to the slope of \( V_0 \) vs \( \nu \) line. 
(D) The value of electric charge of electron is not required to determine \( h \) using the \( V_0 \) vs \( \nu \) curve. 
(E) The work function can be estimated without knowing the value of \( h \). 
Choose the correct answer from the options given below:

• A. (C) and (D) only
• B. (D) and (E) only
• C. (A), (B) and (C) only
• D. (A), (C) and (E) only

Hint:

In the photoelectric effect, the linear relationship between stopping potential and frequency is a result of the work function and Planck's constant. The slope of the curve is crucial in determining \( h \).

Answer 1

The Correct Option is C

The photoelectric equation is given by: \[ V_0 = \frac{h\nu}{e} - \phi_0, \] where: - \( V_0 \) is the stopping potential, - \( \nu \) is the frequency of incident light, - \( h \) is Planck's constant, - \( e \) is the charge of the electron, - \( \phi_0 \) is the work function. From this equation, we can see that \( V_0 \) is linear with respect to \( \nu \), with a slope of \( \frac{h}{e} \), and the intercept gives the value of \( \phi_0 \).

• (A) is true because the relation between \( V_0 \) and \( \nu \) is linear.
• (B) is also true because the slope of the line gives \( \frac{h}{e} \), and rearranging gives \( \frac{\phi_0}{h} \).
• (C) is true because the slope of the \( V_0 \) vs \( \nu \) line is related to \( h \).
• (D) is false because the value of the electric charge \( e \) is required to find \( h \) from the slope.
• (E) is false because to determine the work function, knowing \( h \) is essential, and the value of \( h \) cannot be determined from the \( V_0 \) vs \( \nu \) curve alone without knowing \( e \).

Final Answer: (3) (A), (B) and (C) only.

Answer 2

Step 1: Understand the photoelectric effect and the stopping potential.
In the photoelectric effect, electrons are ejected from a metal surface when light of frequency \( \nu \) strikes it. The stopping potential \( V_0 \) is the potential required to stop the fastest ejected electrons. According to Einstein’s photoelectric equation: \[ E_{\text{kin}} = h\nu - \phi_0, \] where \( h \) is the Planck’s constant, \( \nu \) is the frequency of the incident light, and \( \phi_0 \) is the work function of the metal.
The stopping potential \( V_0 \) is related to the kinetic energy of the ejected electrons: \[ E_{\text{kin}} = eV_0, \] where \( e \) is the charge of the electron.
Thus, the relationship between \( V_0 \) and \( \nu \) can be derived as: \[ eV_0 = h\nu - \phi_0. \] This equation shows that \( V_0 \) is a linear function of \( \nu \), with a slope of \( \frac{h}{e} \).

Step 2: Analyze the options.
- (A) \( V_0 \) vs \( \nu \) is linear.
This is true, as derived above, the stopping potential \( V_0 \) is linearly proportional to the frequency \( \nu \) of the incident light.

- (B) The slope of \( V_0 \) vs \( \nu \) curve is \( \frac{\phi_0}{h} \).
This is incorrect. The slope of the \( V_0 \) vs \( \nu \) curve is actually \( \frac{h}{e} \), not \( \frac{\phi_0}{h} \).

- (C) \( h \) constant is related to the slope of \( V_0 \) vs \( \nu \) line.
This is true. The Planck’s constant \( h \) is related to the slope of the \( V_0 \) vs \( \nu \) curve, as the slope is \( \frac{h}{e} \).

- (D) The value of electric charge of electron is not required to determine \( h \) using the \( V_0 \) vs \( \nu \) curve.
This is incorrect. The charge of the electron \( e \) is required to determine \( h \) using the slope of the curve, as the slope is \( \frac{h}{e} \).

- (E) The work function can be estimated without knowing the value of \( h \).
This is incorrect. The work function \( \phi_0 \) can be estimated only if \( h \) is known, as it is related to the intercept of the \( V_0 \) vs \( \nu \) line by the equation \( \phi_0 = h\nu - eV_0 \).

Step 3: Conclusion.
The correct answers are (A), (B), and (C).

Final Answer:
\[ \boxed{(A), (B) \text{ and } (C) \text{ only}}. \]