Question

An electron in a picture tube of TV set is accelerated from rest through a potential difference of $5\times {{10}^{3}}V$. Then, the speed of electron as a result of acceleration is going to be:
$\begin{aligned} & (A)1.2\times {{10}^{7}}m{{s}^{-1}} \\\ & (B)2.2\times {{10}^{7}}m{{s}^{-1}} \\\ & (C)3.2\times {{10}^{7}}m{{s}^{-1}} \\\ & (D)4.2\times {{10}^{7}}m{{s}^{-1}} \\\ \end{aligned}$

Answer 1

The electron is initially at rest, so its initial kinetic energy is zero. Now, after applying a potential difference, the electron will accelerate and finally attain a constant steady speed. This condition will arrive when all of the potential energy at the start of motion has been converted into kinetic energy of the electron. Thus, we can use the principle of conservation of energy to get the required speed of an electron.

Complete step-by-step answer:
We will first assign some terms which are to be used later in our equations and calculations.
Let the potential difference applied on the electron be given by E .Then, the value of E is given to us in the problem as:
$\begin{aligned} & \Rightarrow E=5\times {{10}^{3}}V \\\ & \therefore E=5000V \\\ \end{aligned}$
Now, let the charge on the electron be given by q .Then, the value of q is known to us as:
$\Rightarrow q=1.6\times {{10}^{-19}}C$
Also, let the mass of one electron be given by m .Then, the value of m is known to us as:
$\Rightarrow m=9.1\times {{10}^{-31}}kg$
And lastly, let the velocity of the electron be given by ‘v’ .Then, the value of ‘v’ is to be calculated. This can be done by energy conservation of electrons. That is, the total potential energy in the start is equal to the steady state (no acceleration) kinetic energy of the electron. Mathematically, this could be written as:
$\Rightarrow qE=\dfrac{1}{2}m{{v}^{2}}$
Putting the values of all the known terms in the above equation, we get:
$\begin{aligned} & \Rightarrow 1.6\times {{10}^{-19}}\times 5000=\dfrac{1}{2}\times 9.1\times {{10}^{-31}}\times {{v}^{2}} \\\ & \Rightarrow {{v}^{2}}=\dfrac{160\times {{10}^{14}}}{9.1} \\\ & \Rightarrow v=\sqrt{\dfrac{160\times {{10}^{14}}}{9.1}}m{{s}^{-1}} \\\ & \Rightarrow v=\sqrt{\dfrac{160}{9.1}}\times {{10}^{7}}m{{s}^{-1}} \\\ & \Rightarrow v=4.192\times {{10}^{7}}m{{s}^{-1}} \\\ & \therefore v\approx 4.2\times {{10}^{7}}m{{s}^{-1}} \\\ \end{aligned}$
Hence, the final speed of the electron as a result of acceleration comes out to be $4.2\times {{10}^{7}}m{{s}^{-1}}$.

So, the correct answer is “Option (D)”.

Note: Application of conservation of energy is not only valid for large measurable objects, but also for very small piece of substance at quantum level like the atoms or electrons. Also, in problems like the one solved above, we should be careful while performing the calculations as they can be long and thus have increased chances of concurring mistakes.