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Which one of the following is true in photoelectric emission
A. Photoelectric current is directly proportional to the amplitude of light of given frequency
B. Photoelectric current is directly proportional to the intensity of light of given frequency at moderate intensities.
C. Above the threshold frequency, the maximum kinetic energy of the photoelectrons is inversely proportional to the frequency of incident light.
D. The threshold frequency depends on the intensity of the incident light.

Answer
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Hint: The minimum amount of energy required for electron emission from the metal surface is the work function of that metal and the frequency of light corresponding to this minimum energy is called threshold frequency and the corresponding wavelength is called threshold wavelength. From the photoelectric effect, if the number of photons hitting the metal surface is higher, more energy transfer will take place leading to an increase in the number of electron emissions which increases the current production.

Formula(e) used:
Energy, \[E = h\nu = \dfrac{{hc}}{\lambda }\]
Equation for photoelectric effect, \[h\nu = {\phi _o} + {E_k}\]
Work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}}\]
Where,
E = energy of incident radiation
\[{\phi _o}\]= Work function of the metal
\[{E_k}\]= energy of the emitted photoelectron
\[c{\rm{ }} = \text{speed of light}= 3 \times {10^8}m/s\]
\[\nu \]= frequency of the light,
\[{\nu _o}\]= threshold frequency,
\[\lambda \]= wavelength of the light,
\[{\lambda _o}\]= threshold wavelength

Complete step by step solution:
A light beam may be thought of as a train of energetic but massless particles known as photons. When a light beam contacts the surface of a metal, photons transmit their energy in quanta (i.e. energy of each photon equal to (\[h\nu \]) to atoms within the metal. The electrons in the atom's ground state obtain this energy as kinetic energy and are effectively expelled from the metal atom. This flow of electrons results in the formation of electric current. This is known as the photoelectric effect.

Einstein discovered it in 1905, for which he also received the Nobel Prize in Physics. The photoelectric effect occurs only if the energy is greater than the metal's threshold energy. These released electrons are known as photoelectrons since their energy is less than that of the incoming light because some of the energy is used to overcome the barrier energy or the work function. Thus, the work function may be defined as the minimal energy necessary for photoelectron emission from a certain metal's surface.

Equation for photoelectric effect is,
\[h\nu = {\phi _o} + {E_k}\]---- (1)
Work function,
\[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}}\]---(2)
For a light of given frequency, photoelectric current is directly proportional to the square of its amplitude. From equation (1), the maximum kinetic energy of the photoelectron is,
\[{E_k} = h\nu - {\phi _o}\]
It is not inversely proportional to the frequency of incident light. Work function, threshold frequency and threshold wavelength are the characteristic intrinsic property of the material and independent of the properties of the incident radiation. Hence work function does not vary with intensity of the incident radiation. Therefore, current is directly proportional to the intensity of the incident radiation.

Hence option B is the correct answer.

Note: For a given frequency of light incident on the metal surface, photoelectric current increases with intensity as more intensity means increased number of photons which in turn causes increase in number of photoelectrons. But this is applicable only if the frequency of incident radiation is greater than or equal to the threshold frequency of that metal.