Question

Light of wavelength \[\lambda \] falls on metal having work function \[\dfrac{{{\text{hc}}}}{{{\lambda _o}}}\] . Photoelectric effect will take place only if:
A.\[\lambda \geqslant {\lambda _o}\]
B.\[\lambda \geqslant {\nu _o}\]
C.\[\lambda \leqslant {\lambda _o}\]
D.\[\lambda \leqslant \dfrac{{{\lambda _o}}}{2}\]

Answer 1

Hint:Energy is directly proportional to frequency but inversely proportional to wavelength. For a successful photoelectric effect, incident light must have energy equal or greater than threshold frequency.

Complete step by step answer:
Photoelectric effect was discovered by Hertz. This effect was explained by Einstein by Quantum theory of light. In this experiment, electrons are ejected from the surface of metal when a beam of light strikes on the surface of metal. The number of electrons ejected is directly proportional to the intensity of the striking light. For each metal, there is a characteristic minimum frequency \[{\nu _o}\] below which the photoelectric effect will not occur. This minimum frequency is also known as threshold frequency. When a light of frequency higher than threshold frequency \[\nu > {\nu _o}\] falls on a metal surface, then the ejected electrons come out with a certain amount of kinetic energy. Kinetic energy of ejected electrons is directly proportional to frequency of striking light. Kinetic energy does not depend on intensity of light.
If the striking light has energy \[{\text{h}}\nu \]or \[\dfrac{{{\text{hc}}}}{\lambda }\] where \[\nu \]the frequency of incident is light and \[\lambda \] is the wavelength of light. The minimum energy required to eject the electron without kinetic energy is given by \[{\text{h}}{\nu _o}\] or \[\dfrac{{{\text{hc}}}}{{{\lambda _o}}}\] where \[{\nu _o}\] is minimum or threshold frequency and \[{\lambda _o}\] is threshold wavelength. Then kinetic energy of ejected electron can be given as:
\[{\text{KE}} = {\text{h}}\nu - {\text{h}}{\nu _o} = \dfrac{{{\text{hc}}}}{\lambda } - \dfrac{{{\text{hc}}}}{{{\lambda _o}}}\] as we already know \[{\text{c}} = \nu \lambda \] .
So, for photoelectric effect to take place light must have energy equal to or greater than threshold frequency or we can say light must have wavelength equal to less than threshold wavelength as frequency is inversely proportional to wavelength.

Thus, the correct option is C.

Note:
When a photon of sufficient energy strikes on metal, the ejection of electrons occurs without time lag. Greater the energy possessed by photons, greater will be kinetic energy as the energy minimum required or work function is constant for a given metal.