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The potential difference between cathode and anode in a cathode ray tube is V. The speed attained by the electrons is proportional to which among the following(a) $V$(b) $\sqrt V$ (c) ${V^2}$ (d) $\dfrac{1}{{\sqrt V }}$

Last updated date: 28th Feb 2024
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Hint: This question can be resolved by determining the kinetic energy of the electron in motion and the electrical potential energy caused by the acceleration of electrons from the negatively charged cathode to the positively charged anode. After this by applying the law of conservation of energy the problem can be solved. Once the solution is obtained, compare it to the options given and obtain the answer.

Complete step-by-step solution:
In a cathode ray tube, electrons emitted from the cathode (negatively charged) move towards the anode (positively charged)
As a result of the electron being in motion, they are associated with Kinetic energy.
Which is given by, Kinetic energy=$\dfrac{1}{2}m{v^2}$ where m- mass of electron; v- velocity of the electron
Also the potential difference (developed because of the presence of cathode and anode and which aids the movement of electrons from cathode to anode) is given by E=eV
where E- Electrical potential energy; e- charge of one electron; V- potential energy
From the law of conservation of energy
We get, Kinetic energy= electrical potential energy
Substituting the values, $\dfrac{1}{2}m{v^2}$=eV
Rearranging we get, ${v^2} = \dfrac{{2eV}}{m}$
The values 2, e and m are constant
Hence we can write, ${v^2} = (\dfrac{{2e}}{m})*V = k*V$
Simplifying we get, $v = \sqrt {k*V}$
Option (b) is the correct answer.

Note: According to the law of conservation of energy, in an isolated system the total energy of the system will remain constant. In numerical problems for these types of equations, care must be taken while mentioning the units of energy, as for tiny particles such as electrons the unit mentioned is eV and not Joules.