Question

The threshold wavelength for photoelectric emission from a material is $520\;nm$. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a:
(multiple options may be right)
A. $50\;W$ infrared lamp
B. $1\;W$ infrared lamp
C. $50\;W$ ultraviolet lamp
D. $1\;W$ ultraviolet lamp

Answer 1

Hint: Recall that in photoelectric effect, photoelectrons are only emitted if the incident light is above a threshold frequency or below a threshold wavelength. Using this as an anchor, determine the relation between the kinetic energy gained by the photoelectrons and the incident and threshold energies, and deduce the relation between the wavelength of incident light and threshold wavelength. Remember that photoelectrons always have finite kinetic energy, without which they cannot escape from the metal surface.

Formula used: Kinetic energy gained by the electrons is given by $KE_{electron} = E_{photon} – E_{threshold} = h(\nu - \nu_0)$, where, $\nu$ is the frequency of incident light and $\nu_{0}$ is the threshold frequency of the metal.

Complete step by step answer:
Let us begin by understanding what photoelectric emission is.
It is a process by which free electrons are emitted from the surface of metals by absorbing electromagnetic radiation (photons) of a suitable frequency that is incident on it. Electrons emitted in this way are called photoelectrons.
Photoelectrons are emitted only when the frequency (energy) of the incident light is greater than the minimum frequency (energy) for photoelectric emission, called the threshold frequency (energy).
Therefore, the electrons on the surface of metals absorb the incident photons and get emitted out by an energetic transfer, i.e., electrons on the surface gain some energy from the photons, and then get emitted with some kinetic energy obtained from the remaining incident energy.
That is, if the energy of the incoming photon is $E_{photon} = h\nu$ where $\nu$ is the frequency of incident light, and,
If the minimum energy required by the metal to emit electrons is given by $E_{threshold} = h\nu_{0}$, where $\nu_{0}$ is the threshold frequency of the metal,
Then the Kinetic energy gained by the electrons is given by $KE_{electron} = E_{photon} – E_{threshold} = h \nu – h\nu_{0} = h(\nu - \nu_0) $.
The above equation can be expressed in terms of wavelength using the relation $\nu = \dfrac{c}{\lambda}$, where c is the light velocity and $\lambda$ is the wavelength of incident light:
$KE_{electron} = hc \left(\dfrac{1}{\lambda} - \dfrac{1}{\lambda_0}\right)$
Now, we know that for an electron to be emitted from the surface of the metal it has to possess some kinetic energy that it can gain from incident energy. Therefore, for a photoelectron:
$KE >0$
$\Rightarrow hc \left(\dfrac{1}{\lambda} - \dfrac{1}{\lambda_0}\right) > 0$
$\Rightarrow \dfrac{1}{\lambda} > \dfrac{1}{\lambda_0}$
$\Rightarrow \lambda < \lambda_0$
And from what is given in the question,
$\Rightarrow \lambda < 520\;nm$
Therefore we have deduced that photoelectrons will be emitted from the material when the light incident from it is less than $520\;nm$,
Now, if we consider the electromagnetic spectrum, we have visible light between $400\;nm$ and $700\;nm$, and below $400\;nm$ we have UV radiation and above $700$ we have the IR radiation.

Thus, from the options given to us the correct choices would be two: C. $50\;W$ UV lamp and D. $1\;W$ UV lamp.

Note: The question poses options that are expressed in terms of power unit watts (W). Note that the magnitude of power has nothing to do with the emitted photoelectrons. All that we are concerned about is the wavelength range of light, as in, UV or IR in this case. The power just contributes to the intensity of the incident light and does not influence the frequency of incident light, which is what photoemission depends on anyways.
Also remember if the incident light is below the threshold frequency, it does not emit electrons, however intense it may be.