Question

The stopping potential (\(V_0\)) versus frequency (\(\nu\)) of a graph for the photoelectric effect in a metal is given. From the graph, the Planck's constant (\(h\)) is: 

• A. \(6.60 \times 10^{-34} \, {J·s}\)
• B. \(6.69 \times 10^{-34} \, {J·s}\)
• C. \(6.62 \times 10^{-34} \, {J·s}\)
• D. \(6.63 \times 10^{-34} \, {J·s}\)

Hint:

For graph-based problems in the photoelectric effect, remember: - The slope of the \(V_0\) vs. frequency graph gives \( h/e \). - Multiply by \( e \) to get \( h \).

Answer 1

The Correct Option is A

Step 1: Understanding the Photoelectric Equation The photoelectric equation is given by: \[ eV_0 = h(\nu - \nu_0) \] where: \( eV_0 \) is the stopping potential energy, \( h \) is Planck’s constant, \( \nu \) is the frequency of incident light, \( \nu_0 \) is the threshold frequency. 
Step 2: Determining the Slope from the Graph The given graph shows a straight-line relationship between \(V_0\) and frequency \(\nu\), where the slope represents: \[ {Slope} = \frac{h}{e} \] From the graph: \[ {Slope} = \frac{16.5 { V}}{(8 - 4) \times 10^{14} { Hz}} = \frac{16.5}{4 \times 10^{14}} \] \[ = 4.125 \times 10^{-14} { V·s} \] 
Step 3: Calculating Planck’s Constant Since \( e = 1.6 \times 10^{-19} { C} \), we calculate \( h \) as: \[ h = ({Slope}) \times e \] \[ = (4.125 \times 10^{-14}) \times (1.6 \times 10^{-19}) \] \[ = 6.60 \times 10^{-34} { J·s} \]