Dual Nature of Matter and Radiation Test - 4

The maximum kinetic energy for the photoelectrons is  
E_max ​=h ν−ϕ
where, νis the frequency of incident light and ϕis photoelectric work function of metal.
If V_o ​is the stopping potential then
eV₀ ​=h(c/λ)​−ϕ.....................(since, ν=c/λ​)
As per the problem, for incident light of 4000A^o , the stopping potential is 2V. When the wavelength of incident light is reduced to 3000A^o , then the stopping potential will increase to value more than 2V(as per the above equation).

The number of electrons emitted per second is observed to be directly proportional to the intensity of light. “Ok, so light is a wave and has energy. It takes electrons out of a metal, what is so special about that!”First of all, when the intensity of light is increased, we should see an increase in the photocurrent (number of photoelectrons emitted). Right?
As we see, this only happens above a specific value of frequency, known as the threshold frequency. Below this threshold frequency, the intensity of light has no effect on the photocurrent! In fact, there is no photocurrent at all, however high the intensity of light is.

The graph between the photoelectric current and the intensity of light is a straight line when the frequency of light used is above a specific minimum threshold value.

Photoelectric effect can be explained by assuming light consisting of quanta that when a photon of certain energy falls on the surface of metal, then the metal surface starts to emit an electron known as photo electrons.

The Stopping potential =2.5V.
or, Kinetic energy=2.5eV.
We know that,
Incident energy =work function + Kinetic energy.
To get incident energy in e.V,
We also know that,
The Incident energy =12400/λÅ
Incident energy=work function + kinetic energy.
12400/3000 = work function + 2.5e.v.
4.13-2.5 = work function
work function=1.64 e.V

When intensity of incident photons increases, the number of electrons emitted from the surface increases due to which current increases. Frequency of incident photons only limits the maximum kinetic energy of the photoelectron emitted and so the current is independent of frequency of photon.

Photoelectric current is zero when the stopping potential is sufficient to repel even the most energetic photoelectrons with the maximum kinetic energy K_max ​so that K_max ​=eV_{0 ​}
For a given frequency of the incident radiation, the stopping potential is independent of its intensity.

Relation between kinetic energy and frequency is K.E+ϕ=h ν.
Here h is planck 's constant, ϕ=constant (Work function)
According to the above equation of photoelectric effect, as we increase frequency of photon, kinetic energy of photoelectron increases. Because the frequency of incident light is directly proportional to kinetic energy.
Hence option C is correct.
 

A photon is a particle of light which essentially is a packet of electromagnetic radiation. The energy of the photon depends on its frequency (how fast the electric field and magnetic field wiggle). The higher the frequency, the more energy the photon has. Of course, a beam of light has many photons. This means that really intense red light (lots of photons, with slightly lower energy) can carry more power to a given area than less intense blue light (fewer photons with higher energy).
 The speed of light (c) in a vacuum is constant. This means more energetic (high frequency) photons like X-rays and gamma rays travel at exactly the same speed as lower energy (low frequency) photons, like those in the infrared. As the frequency of a photon goes up, the wavelength goes down, and as the frequency goes down, the wavelength increases.
 

The number of electrons ejected can be measured as a function of intensity.
Intensity is equal to energy per unit time per unit area.
Keeping frequency constant, increasing intensity will increase energy thus increase of ejected electrons.
Thus,
Number of ejected photoelectrons will increase with increase in intensity of light.