Question

An $\alpha$-particle of mass $6.4\times10^{-27}$ kg and charge $3.2\times10^{-19}$ C is situated in a uniform electric field of $1.6\times10^{5}$ V m$^{-1}$. The velocity of the particle at the end of $2\times10^{-2}$ m path when it starts from rest is

• A. $2\sqrt{3}\times10^5~\text{ms}^{-1}$
• B. $8\times10^5~\text{ms}^{-1}$
• C. $16\times10^5~\text{ms}^{-1}$
• D. $4\sqrt{2}\times10^5~\text{ms}^{-1}$

Hint:

Combine $F=qE$ and $v^2=2as$ for charged particles in electric fields.

Answer 1

The Correct Option is D

Step 1: Acceleration
Force $F = qE = 3.2\times10^{-19} \times 1.6\times10^5 = 5.12\times10^{-14}~\text{N}$. Acceleration $a = \frac{F}{m} = \frac{5.12\times10^{-14}}{6.4\times10^{-27}} = 0.8\times10^{13}~\text{ms}^{-2}$.
Step 2: Kinematics
Using $v^2 = u^2 + 2as$ with $u=0$.
Step 3: Calculation
$v^2 = 2 \times 0.8\times10^{13} \times 2\times10^{-2} = 3.2\times10^{11} = 32\times10^{10}$. Thus, $v = \sqrt{32\times10^{10}} = 4\sqrt{2}\times10^5~\text{ms}^{-1}$.
Final Answer: (d)