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

Concept: The photoelectric effect describes the emission of electrons from a metal surface when light of sufficient frequency falls on it. According to Einstein’s photoelectric equation, the maximum kinetic energy of emitted electrons depends on the frequency of incident radiation. The equation is given by: \[ K.E._{\text{max}} = h\nu - \phi \] where \(K.E._{\text{max}}\) = maximum kinetic energy of emitted electrons \(h\) = Planck’s constant \(\nu\) = frequency of incident radiation \(\phi\) = work function of the metal

Step 1: Relation between frequency and wavelength. The frequency of light is related to wavelength by: \[ \nu = \frac{c}{\lambda} \] where \(c\) = speed of light \(\lambda\) = wavelength of incident radiation

Step 2: Substitute frequency in Einstein’s equation. \[ K.E._{\text{max}} = h\left(\frac{c}{\lambda}\right) - \phi \]

Step 3: Determine proportionality. For a given metal, \(h\), \(c\), and \(\phi\) are constants. Therefore, the maximum kinetic energy mainly depends on \(\frac{1}{\lambda}\). \[ K.E._{\text{max}} \propto \frac{1}{\lambda} \] Thus, as wavelength decreases, the maximum kinetic energy of emitted electrons increases.