Hydrogen ion and singly ionised helium atom are accelerated from rest, through the same potential di

Q: Why is the charge of singly ionized helium $e$ and not $2e$?

A neutral helium atom has 2 protons in its nucleus and 2 electrons orbiting it. 'Singly ionized' means it has lost exactly one electron. Therefore, it has 2 positive charges and 1 negative charge, leaving a net charge of $+1e$.

Q: Why do we assume the mass of helium is $4m$?

A hydrogen ion ($H^+$) is just a single proton, so its mass is $m_p$. A helium nucleus contains 2 protons and 2 neutrons. Since protons and neutrons have roughly the same mass, the total mass of the helium ion is approximately $4m_p$. The mass of the remaining electron is negligible.

Q: Does the distance between the plates matter?

No, the distance doesn't matter for the final velocity. The work done by the electric field depends only on the total potential difference $V$ the particle moves through, not the distance over which that potential drops.

Q: What if they didn't start from rest?

If they had an initial velocity $u$, the work-energy theorem would be $qV = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$. The final velocity would then depend on both the potential difference and their initial speeds.

Q: How would the answer change if it was an alpha particle?

An alpha particle is a doubly ionized helium atom ($He^{2+}$). Its charge would be $2e$ and its mass $4m$. The velocity would be $v_\alpha = \sqrt{2(2e)V / 4m} = \sqrt{eV/m}$. The ratio $v_H / v_\alpha$ would then be $\sqrt{2} : 1$.

Question

Hydrogen ion and singly ionised helium atom are accelerated from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to

Options:

(A) 1 : 2

(B) 10 : 7

(C) 2 : 1

(D) 5 : 7

Sigma Percentile · Accepted Answer

### The Setup: Accelerating through a Potential Difference

Imagine a classic physics setup: two parallel metal plates connected to a battery, creating a constant potential difference $V$ between them. In this electric field, we place two tiny contestants at the positive plate, both starting from rest.

Our first contestant is a **hydrogen ion** ($H^+$), which is essentially a bare proton. Our second contestant is a **singly ionized helium atom** ($He^+$). The electric field exerts a force on both, accelerating them towards the negative plate. The question is: who crosses the finish line faster, and by what ratio?

### The Master Equation: Energy Conservation

To solve this, we don't need to track their complex kinematics step-by-step. Instead, we use the elegant principle of **Energy Conservation**.

When a charged particle of charge $q$ moves through a potential difference $V$, the electric field does work on it. This work is entirely converted into the particle's kinetic energy. Mathematically, we write this as:

$$qV = \frac{1}{2}mv^2$$

Since both particles start from rest, their initial kinetic energy is zero. We can rearrange this master equation to find an expression for the final velocity $v$:

$$v = \sqrt{\frac{2qV}{m}}$$

Notice that the potential difference $V$ is a constant for both ions. The final velocity depends solely on the ratio of their charge to their mass, often called the specific charge.

### Comparing the Contenders: Hydrogen vs. Helium

Let's analyze our two contestants based on our velocity formula.

**For the Hydrogen Ion ($H^+$):**
It has a charge of $+1e$ (where $e$ is the elementary charge) and a mass we'll call $m$. Plugging these into our formula, its final velocity $v_H$ is:

$$v_H = \sqrt{\frac{2eV}{m}}$$

**For the Singly Ionized Helium Atom ($He^+$):**
"Singly ionized" means it has lost exactly one electron. Therefore, its net charge is also $+1e$. However, a helium nucleus contains two protons and two neutrons, making its mass approximately four times that of a hydrogen ion. So, its mass is $4m$. Its final velocity $v_{He}$ is:

$$v_{He} = \sqrt{\frac{2eV}{4m}}$$

### The Final Showdown: Calculating the Ratio

Now for the grand finale. We need the ratio of their final speeds, $v_H : v_{He}$. Let's divide the two expressions we just found:

$$\frac{v_H}{v_{He}} = \frac{\sqrt{\frac{2eV}{m}}}{\sqrt{\frac{2eV}{4m}}}$$

Because they were accelerated through the same potential difference $V$, and they have the same charge $e$, the terms $2eV$ cancel out beautifully. The mass $m$ also cancels out, leaving us with:

$$\frac{v_H}{v_{He}} = \sqrt{\frac{4m}{m}} = \sqrt{4} = 2$$

This means the hydrogen ion is exactly twice as fast as the singly ionized helium atom when they reach the other plate. The final ratio is **$2 : 1$**.

Animated Solution for Physics - Electrostatics: Hydrogen ion and singly ionised helium atom are accelerated from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to

Select Answer:

Visualized Solution

Visualizing the Setup

Energy Conservation Principle

Velocity Expression

Velocity of Hydrogen Ion

Velocity of Helium Ion

Calculating the Ratio

Food for Thought

The Sigma Insight: Electric Potential and Potential Difference

The Setup

The Master Equation

Comparing the Contenders

The Final Showdown

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