Question

What is the relationship between maximum kinetic energy and wavelength in the photoelectric effect?

• A. \(K.E. \propto \lambda\)
• B. \(K.E. \propto \lambda^2\)
• C. \(K.E. \propto \frac{1}{\lambda}\)
• D. \(K.E. \propto \sqrt{\lambda}\)

Hint:

In the photoelectric effect, higher frequency (or smaller wavelength) light gives electrons more kinetic energy. Hence, \(K.E._{\text{max}} \propto \frac{1}{\lambda}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks how the maximum kinetic energy (\(K.E._{max}\)) of photoelectrons relates to the wavelength (\(\lambda\)) of the incident light.
Step 2: Key Formula or Approach:
Einstein’s photoelectric equation relates kinetic energy to incident frequency (\(\nu\)):
\[ K.E._{max} = h\nu - \phi \]
where \(h\) is Planck's constant and \(\phi\) is the work function.
Step 3: Detailed Explanation:
The frequency \(\nu\) is related to wavelength \(\lambda\) by the formula \(\nu = \frac{c}{\lambda}\), where \(c\) is the speed of light.
Substituting this into the Einstein equation:
\[ K.E._{max} = \frac{hc}{\lambda} - \phi \]
In this expression, \(hc\) and \(\phi\) are constants for a specific metal and light source. The term \(\frac{hc}{\lambda}\) shows that as the wavelength \(\lambda\) increases, the energy of the incident photon decreases, and consequently, the kinetic energy decreases.
Thus, \(K.E._{max}\) is inversely proportional to the wavelength.
Step 4: Final Answer:
The relationship is \(K.E. \propto \frac{1}{\lambda}\).