Question

In an experiment to study photo-electric effect the observed variation of stopping potential with frequency of incident radiation is as shown in the figure. The slope and y-intercept are respectively

• A. \(\frac{hv}{e},v_0\)
• B. \(\frac{hv}{e},-\frac{h}{e}\)
• C. hv, -hv₀
• D. \(\frac{h}{e},-\frac{hv_0}{e}\)

Answer 1

The Correct Option is D

Step 1: Recall Einstein’s photoelectric equation clearly:

The photoelectric equation is given by: 

\[ eV_0 = h\nu - h\nu_0 \]

where:
\( e \) = electronic charge,
\( V_0 \) = stopping potential,
\( h \) = Planck’s constant,
\( \nu \) = frequency of incident radiation,
\( \nu_0 \) = threshold frequency.

Step 2: Rewrite the equation in linear form (\(y=mx+c\)):

Expressing \(V_0\) clearly in terms of frequency \(\nu\):

\[ V_0 = \frac{h}{e}\nu - \frac{h\nu_0}{e} \]

This equation is now in the standard linear form:

\[ y = mx + c \]

where:
\(y = V_0\),
\(x = \nu\),
\(m = \frac{h}{e}\) (slope),
\(c = -\frac{h\nu_0}{e}\) (y-intercept).

Step 3: Clearly identify slope and y-intercept from above comparison:

• Slope (\(m\)): \(\frac{h}{e}\)
• Y-intercept (\(c\)): \(-\frac{h\nu_0}{e}\)

Final Conclusion:
Slope: \(\frac{h}{e}\), Y-intercept: \(-\frac{h\nu_0}{e}\).

Answer 2

In the photoelectric effect experiment, the stopping potential \( V_0 \) is plotted against the frequency \( v \) of incident radiation. The stopping potential is related to the frequency of the incident radiation by the equation: \[ V_0 = \frac{h}{e} \left( v - v_0 \right) \] where:
\( h \) is Planck's constant,
\( e \) is the charge of the electron,
\( v_0 \) is the threshold frequency, and
\( v \) is the frequency of the incident radiation.

The slope of the graph is \( \frac{h}{e} \), and the y-intercept represents the value \( - \frac{h v_0}{e} \), corresponding to the threshold frequency.

Thus, the correct option is \( \frac{h}{e} ,- \frac{h v_0}{e} \).