Question

How does the maximum kinetic energy of photoelectrons relate to wavelength (\(\lambda\))?

• 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:

Shorter wavelength light has higher photon energy, therefore it produces photoelectrons with greater kinetic energy. Hence \(K.E \propto \frac{1}{\lambda}\).

Answer 1

The Correct Option is C

Concept: The photoelectric effect is the phenomenon in which electrons are emitted from the surface of a metal when light of sufficient frequency falls on it. This effect provided strong evidence for the particle nature of light and was explained by Albert Einstein using the concept of photons. When light strikes a metal surface, each photon carries energy given by: \[ E = h\nu \] where \(h\) = Planck's constant \(\nu\) = frequency of the incident light. Since frequency and wavelength are related by the speed of light \(c\), \[ \nu = \frac{c}{\lambda} \] Substituting this into the energy expression: \[ E = \frac{hc}{\lambda} \] According to Einstein’s photoelectric equation: \[ K_{\text{max}} = h\nu - \phi \] where \(K_{\text{max}}\) = maximum kinetic energy of emitted electrons \(\phi\) = work function of the metal. Substituting \( \nu = \frac{c}{\lambda} \): \[ K_{\text{max}} = \frac{hc}{\lambda} - \phi \] This shows that the kinetic energy varies inversely with wavelength.

Step 1: Energy of incident photon.
The energy carried by a photon is proportional to the frequency of light.

Step 2: Relation between frequency and wavelength.
Frequency is inversely proportional to wavelength: \[ \nu = \frac{c}{\lambda} \]

Step 3: Dependence of kinetic energy.
Since kinetic energy depends on photon energy, it follows that: \[ K.E \propto \frac{1}{\lambda} \] Thus, the maximum kinetic energy of photoelectrons is inversely proportional to wavelength.