Question

When a charged particle of charge $xC$ moves through a potential difference of $yV$, the gain in kinetic energy is equal to $xyJ$.
An electron and an alpha particle have their masses in the ratio of 1 : 7200 and charges in the ratio of 1 : 2. If they start moving from rest through the same electrical potential difference then the ratio of their velocities will be ______.

• A. 1 : 60
• B. 60 : 1
• C. 1 : 20
• D. 20 : 1

Answer 1

The correct option is B 60 : 1

Given : If particle has charge $x$ coulombs and potential difference $y$ volts . Then gain in K.E. = $xy$ joules.

Using similar analysis,

$\frac{\text{(mass)electron}}{\text{(mass) alpha particle}}=\frac{1}{7200}$

$\frac{\text{(charge)electron}}{\text{(charge) alpha particle}}=\frac{1}{2}$

We need to find $\frac{\text{(velocity)electron}}{\text{(velocity) alpha particle}}$

Since initially there at rest, initial K.E. = 0;

They move through same potential difference. Let potential difference be $y$ volts.

So, gain in K.E = Final K.E – Initial K.E

= Final K.E – 0

So, gain in K.E = $charge\times potentialdifference$

Ratio of gain in K.E of electron and $\alpha$ particle = Ratio of final K.E of electron and $\alpha$ particle

$\frac{\frac{1}{2}m{v}_e^2\times y}{\frac{1}{2}m{v}_{\alpha }^2\times y}$

$=\frac{1}{2}\times 7200=3600$

$\frac{v_e}{v_{\alpha }}=\sqrt{3600}=60$