Question

A proton is released from rest at the positive plate of a parallelplate capacitor. It crosses the capacitor and reaches the negative plate with a speed of $50,000\mathrm{m}/\mathrm{s}$. The experiment is repeated with a${\mathrm{He}}^{+}$ ion (charge $\mathrm{e}$, mass$4\mathrm{u}$ ). What is the ion's speed at the negative plate?

Short answer

Ion's speed at the negative plate is$25,000\mathrm{m}/\mathrm{s}.$

Step-by-step solution

Formula for law of conversion of energy

Given information:

A proton speed in negative plate is $50,000m/s$.

Find:

The objective is to find ion's speed at the negative plate.

According to the law of energy discussion

we obtain,

$\mathrm{q\Delta V}=\frac{{\mathrm{mv}}^2}{2}$

$\mathrm{v}=\sqrt{\frac{2\mathrm{q\Delta V}}{\mathrm{m}}}$

localid="1648838413956" $\mathrm{m}$-mass

Calculation for speed

The experiment is performed with the same potential difference this time.

In addition, the charge remains constant.

As a result, the speed of the ion will be ,

${\mathrm{v}}_2=\sqrt{\frac{2\mathrm{q\Delta V}}{4\mathrm{u}}}$

localid="1648837345219" $=\frac{1}{2}\sqrt{\frac{2\mathrm{q\Delta V}}{\mathrm{u}}}$

localid="1648838082458">=0.52×(1.6×10-19)×50000u

$=25,000\mathrm{m}/\mathrm{s}$