Question

A cathode ray tube is operating at $5kV$. Then what is the K.E acquired by the electrons?
A. $5eV$
B. $5MeV$
C. $5keV$
D. $5V$

Answer 1

Hint:A cathode ray tube experiment shows the occurrence of a phenomenon known as photoelectric effect. The concept of photoelectric effect and the emission of electrons is applied. The formula for the kinetic energy in terms of the stopping potential and the charge of the electron is applied in order to determine the kinetic energy acquired by the electrons

Formula used:
The kinetic energy of the photo-electrons is given by the equation:
$K = e{V_0}$
Where, $e$ denotes the charge of the electron and ${V_0}$ denotes the stopping potential of the electrons.

Complete step by step answer:
The above problem revolves around the concept of photoelectric effect and we are asked to find the kinetic energy of the electrons. In order to determine this, we first need to look into the concept of photoelectric emission inside a cathode ray tube. The phenomenon of emission of electrons from a metal surface when electromagnetic radiations of sufficiently high frequency are incident on it is known as photoelectric effect. Photons or light generated electrons are emitted out once the light is incident on the surface.

When the light is incident on a surface the electrons that are said to eject out of the material are said to acquire or contain some energy which is the kinetic energy gained by the photons due to the incident light on the material. The condition for photoelectric effect to occur is that the frequency of light incident on it should be more than the threshold frequency of the metal or in other words the energy of incident light must have an energy more than that of the work function of the metal.

A cathode ray tube experiment is performed in order to depict the photoelectric effect that occurs due to a light that is incident on the cathode material which tends to emit negatively charged particles which are known to be the photons. These were collected by the anode placed on the other side of the tube and correspondingly current is produced.

Once the saturation point of photo-electric current is attained the photoelectric effect becomes constant and stays at the same value. Thus when a retarding potential is applied to the anode the electrons repel instead of getting collected and hence the concept of stopping potential was introduced.

Now, let us construct an equation relating the stopping potential and the kinetic energy of the ejected electrons. The stopping potential is the minimum value of negative potential applied to the anode of a photocell to make the photoelectric current equivalent to zero. The point where there are no electrons emitted is when the stopping potential is reached. The work done by the stopping potential on the fastest electron trying to escape the surface of the metal will be equivalent to its kinetic energy. Thus the equation for the kinetic energy in terms of the stopping potential for electrons is given by:
$K = e{V_0}$ --------($1$)

Let us now extract the data given in the question. The question says that the cathode tube is operating at a potential of $5kV$ which will be taken to be stopping potential of the photoelectrons. Hence:Given, ${V_0} = 5kV$. Hence the above value is substituted in equation ($1$) to obtain the corresponding kinetic energy of the ejected electrons. Thus we get:
$K = e \times 5kV$
$ \therefore K.E = 5keV$
Hence, the kinetic energy of the ejected electrons has a kinetic energy of value $5keV$.

Therefore, the correct answer is option C.

Note:The photoelectric effect takes place at only certain conditions. Different substances undergo photoelectric effect and emit photons only when exposed to radiations of different frequency and only when it surpasses the minimum required energy needed to emit photons. The photoelectric effect depends on the intensity of light, the potential difference applied between the two electrons and the nature of the cathode material.