Question

A proton, a deuteron and an alpha particle are accelerated through potentials of $V,2V$ and $4V$ respectively. Their velocity will bear a ratio:

• A. $1:1:1$
• B. 1 : $\sqrt{2}$ : 1
• C. $\sqrt{2}$ : 1 : 1
• D. 1 : 1 : $\sqrt{2}$

Answer 1

The correct option is D 1 : 1 : $\sqrt{2}$

Here, kinetic energy $=\frac{1}{2}m{v}^2=eV\Rightarrow v=\sqrt{\frac{2eV}{m}}$
For proton ${(}_1^{1}H):v_p=\sqrt{\frac{2eV}{1}}=\sqrt{2eV}$
For deuteron ${(}_1^{2}H):v_d=\sqrt{\frac{2e\left(2V\right)}{2}}=\sqrt{2eV}$
For alpha ${(}_2^{4}He):v_a=\sqrt{\frac{2\left(2e\right)\left(4V\right)}{4}}=\sqrt{4eV}$
$\therefore v_p:v_d:v_a=1:1:\sqrt{2}$