Question

\(K_1\) and \(K_2\) be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength \(\lambda_1\) and \(\lambda_2\), respectively. If \(\lambda_1 = 2\lambda_2\) then the work function of material is given by:

• A. \(K_2 + 2K_1\)
• B. \(2K_2 - K_1\)
• C. \(K_1 - 2K_2\)
• D. \(K_2 - 2K_1\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The photoelectric effect relates the maximum kinetic energy of emitted electrons to the energy of incident photons and the work function of the material. We set up two equations for the two different wavelengths and solve for the work function.
Step 2: Key Formula or Approach:
Einstein's photoelectric equation: $K_{\text{max}} = \frac{hc}{\lambda} - \Phi$.
Substitute the given relationship $\lambda_1 = 2\lambda_2$ into the equations and eliminate the $\frac{hc}{\lambda}$ term.
Step 3: Detailed Explanation:
For wavelength $\lambda_1$, the kinetic energy is:
$K_1 = \frac{hc}{\lambda_1} - \Phi$ \quad \dots(1)
For wavelength $\lambda_2$, the kinetic energy is:
$K_2 = \frac{hc}{\lambda_2} - \Phi$ \quad \dots(2)
We are given that $\lambda_1 = 2\lambda_2$, which implies $\frac{1}{\lambda_2} = \frac{2}{\lambda_1}$.
Substitute this into equation (2):
$K_2 = \frac{2hc}{\lambda_1} - \Phi$.
From equation (1), we can express $\frac{hc}{\lambda_1}$ in terms of $K_1$ and $\Phi$:
$\frac{hc}{\lambda_1} = K_1 + \Phi$.
Substitute this expression into our modified equation (2):
$K_2 = 2(K_1 + \Phi) - \Phi$.
Expand and simplify:
$K_2 = 2K_1 + 2\Phi - \Phi$.
$K_2 = 2K_1 + \Phi$.
Rearrange to solve for the work function $\Phi$:
$\Phi = K_2 - 2K_1$.
Step 4: Final Answer:
The work function is $K_2 - 2K_1$.