Question

Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference.

• A. 2:1
• B. 1:2
• C. 5:7
• D. 10:7

Answer 1

The correct option is A

2:1

Explanation for the correct option:

The correct option (A)

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{work}\mathrm{energy}\mathrm{theorem},\mathrm{gain}\mathrm{in}\mathrm{kinetic}\mathrm{energy}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{work} \\ \mathrm{done}\mathrm{in}\mathrm{displacement}\mathrm{of}\mathrm{charge}. \\ \mathrm{so}\frac{1}{2}{\mathrm{mv}}^2=\mathrm{q}∆\mathrm{v} \\ \mathrm{Here},∆\mathrm{v}=\mathrm{potential}\mathrm{difference}\mathrm{between}\mathrm{two}\mathrm{positions}\mathrm{of}\mathrm{charges}\mathrm{q}. \\ \mathrm{For}\mathrm{same}\mathrm{q}\mathrm{and}∆\mathrm{v}. \\ \mathrm{v}\propto \frac{1}{\sqrt{\mathrm{m}}} \\ \mathrm{Mass}\mathrm{of}\mathrm{Hydrogen}\mathrm{ion}{\mathrm{m}}_{\mathrm{H}}=1 \\ \mathrm{Mass}\mathrm{ofHelium}\mathrm{ion}{\mathrm{m}}_{\mathrm{He}}=4 \\ \mathrm{so},\frac{{\mathrm{v}}_{\mathrm{H}}}{{\mathrm{v}}_{\mathrm{He}}}=\sqrt{\frac{4}{1}}=2:1 \end{gathered}$$

Final Answer: Hence the correct option is option A.