Question

Kinetic energy of photo electron \(E_K\) changes with frequency \((f)\) of light. Which of the following graph represents this emission?

• A. Graph A
• B. Graph B
• C. Graph C
• D. Graph D

Hint:

For photoelectric effect, remember: \[ E_K=hf-\phi \] So \(E_K\) varies linearly with frequency \(f\), and the slope of the graph is Planck's constant \(h\).

Answer 1

The Correct Option is B

Concept:
This question is based on the photoelectric effect. According to Einstein's photoelectric equation: \[ E_K=hf-\phi \] where, \[ E_K=\text{maximum kinetic energy of photoelectron} \] \[ h=\text{Planck's constant} \] \[ f=\text{frequency of incident light} \] \[ \phi=\text{work function of the metal} \]

Step 1: Understand the relation between kinetic energy and frequency.
The equation is: \[ E_K=hf-\phi \] This is similar to the equation of a straight line: \[ y=mx+c \] Here: \[ y=E_K \] \[ x=f \] \[ m=h \] \[ c=-\phi \] So, the graph of \(E_K\) versus \(f\) is a straight line.

Step 2: Understand the slope of the graph.
In the equation: \[ E_K=hf-\phi \] the coefficient of \(f\) is \(h\). Since Planck's constant \(h\) is positive, the graph must have a positive slope. Therefore, as frequency increases, kinetic energy also increases.

Step 3: Understand threshold frequency.
Photoelectrons are emitted only when the frequency of incident light is greater than or equal to the threshold frequency. At threshold frequency \(f_0\): \[ E_K=0 \] So: \[ hf_0-\phi=0 \] \[ hf_0=\phi \] \[ f_0=\frac{\phi}{h} \] This means the ideal graph cuts the frequency axis at \(f=f_0\).

Step 4: Choose the correct graph from the options.
The graph must be: \[ \text{a straight line with positive slope} \] Among the given options, only Graph B shows a straight line increasing with frequency. Therefore, Graph B represents the relation between kinetic energy of photoelectron and frequency of light. Hence, the correct answer is: \[ \boxed{(B)\ \text{Graph B}} \]