Question

Radiation of wavelength \(\lambda\) is incident on a photosensitive surface. Find the de Broglie wavelength of electrons emitted from the surface. Assume that the work function of the surface is negligible.

Hint:

For photoelectric problems with negligible work function, \[ K=\frac{hc}{\lambda}. \] Then use \[ \lambda_d=\frac{h}{\sqrt{2mK}} \] to obtain the de Broglie wavelength of the emitted electron.

Answer 1

Concept: When electromagnetic radiation falls on a photosensitive surface, electrons may be emitted through the photoelectric effect. According to Einstein's photoelectric equation, \[ h\nu=\phi+K_{\max}. \] If the work function is negligible, \[ \phi \approx 0. \] In that case, the entire photon energy is converted into the kinetic energy of the emitted electron. The de Broglie wavelength of the emitted electron can then be calculated using the kinetic energy obtained from the incident photon.

Step 1: Calculate the energy of the incident photon. For radiation of wavelength \[ \lambda, \] the photon energy is \[ E=\frac{hc}{\lambda}. \] Since the work function is negligible, \[ K=\frac{hc}{\lambda}. \] Thus, the kinetic energy of the emitted electron is \[ \boxed{K=\frac{hc}{\lambda}}. \]

Step 2: Relate kinetic energy and momentum of the electron. For a non-relativistic electron, \[ K=\frac{p^2}{2m}. \] Substituting the value of kinetic energy, \[ \frac{p^2}{2m} = \frac{hc}{\lambda}. \] Multiplying both sides by \(2m\), \[ p^2 = \frac{2mhc}{\lambda}. \] Therefore, \[ p = \sqrt{\frac{2mhc}{\lambda}}. \]

Step 3: Apply de Broglie relation. The de Broglie wavelength of the emitted electron is \[ \lambda_d=\frac{h}{p}. \] Substituting the value of momentum, \[ \lambda_d = \frac{h} {\sqrt{\dfrac{2mhc}{\lambda}}}. \]

Step 4: Simplify the expression. \[ \lambda_d = \sqrt{ \frac{h^2\lambda} {2mhc} }. \] Cancelling one factor of \(h\), \[ \lambda_d = \sqrt{ \frac{h\lambda} {2mc} }. \] Hence, \[ \boxed{ \lambda_d = \sqrt{\frac{h\lambda}{2mc}} }. \]

Step 5: Write the final answer. Therefore, the de Broglie wavelength of the emitted photoelectron is \[ \boxed{ \lambda_d = \sqrt{\frac{h\lambda}{2mc}} }. \]