An alpha-particle and a proton are accelerated from rest by a potential difference of 100 V. After t

Q: Why do we use $K = qV$?

When a charged particle is accelerated from rest through a potential difference $V$, the electrical work done on it ($W = qV$) is entirely converted into its kinetic energy according to the work-energy theorem.

Q: What is an alpha particle exactly?

An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons. Therefore, its charge is $+2e$ (twice that of a proton) and its mass is approximately $4m_p$ (four times that of a proton).

Q: Does the 100 V potential difference matter?

In this specific question, no. Because both particles are accelerated through the *same* potential difference, the $V$ term cancels out when we take the ratio of their wavelengths.

Q: Why is the proton's wavelength longer?

The proton has a smaller mass and a smaller charge compared to the alpha particle. This means it gains less momentum when accelerated through the same potential, and since wavelength is inversely proportional to momentum, the proton ends up with a longer wavelength.

Q: How do we estimate $\sqrt{8}$ without a calculator?

We know that $2^2 = 4$ and $3^2 = 9$. Since 8 is very close to 9, $\sqrt{8}$ must be slightly less than 3. Specifically, it's $2\sqrt{2} \approx 2 \times 1.414 = 2.828$.

Question

An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100	ext{ V}$. After this, their de-Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\frac{\lambda_p}{\lambda_\alpha}$, to the nearest integer, is

Sigma Percentile · Accepted Answer

### The Setup: Racing Particles

Imagine a proton and an alpha particle, both starting from rest, being accelerated through the exact same potential difference of $100	ext{ V}$. As they move between the plates, the electrical work done on them is converted entirely into kinetic energy.

Because they are in motion, they exhibit wave-like properties according to quantum mechanics. Our goal is to find the ratio of their de Broglie wavelengths.

### The Master Equation: de Broglie Wavelength

The de Broglie wavelength $\lambda$ is given by Planck's constant $h$ divided by the particle's momentum $p$:

$$\lambda = \frac{h}{p}$$

We can express momentum in terms of kinetic energy $K$, giving us $p = \sqrt{2mK}$. Since the particles are accelerated by a potential $V$, their kinetic energy $K$ is simply their charge $q$ times $V$. Substituting this back into our wavelength equation, we get the master equation for this problem:

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

### The Ratio: Canceling the Constants

We need to find the ratio of their wavelengths, $\frac{\lambda_p}{\lambda_\alpha}$. Notice that Planck's constant $h$, the number $2$, and the potential difference $V$ are the exact same for both particles. This means the wavelength is inversely proportional to the square root of the product of mass and charge:

$$\lambda \propto \frac{1}{\sqrt{mq}}$$

Therefore, the ratio $\frac{\lambda_p}{\lambda_\alpha}$ is equal to the square root of the mass of the alpha particle times its charge, divided by the mass of the proton times its charge:

$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$$

### The Final Calculation: Nearest Integer

Let's plug in what we know about these particles. An alpha particle is a helium nucleus, consisting of two protons and two neutrons. This makes its mass approximately four times the mass of a proton ($m_\alpha = 4m_p$), and its charge exactly twice the charge of a proton ($q_\alpha = 2q_p$).

Substituting these values into our ratio equation, the masses and charges cancel out beautifully:

$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{(4m_p)(2q_p)}{m_p q_p}} = \sqrt{8}$$

The square root of $8$ is $2\sqrt{2}$, which is approximately $2.828$. The question asks for the nearest integer. Rounding $2.828$ gives us **$3$**.

Interestingly, the $100	ext{ V}$ mentioned in the problem was actually extra information! Because both particles were accelerated through the same potential, the $V$ terms canceled out completely. The ratio of their wavelengths is a constant, regardless of the voltage used.

Enter Numerical Value:

Visualized Solution

$The Setup$

$de-Broglie Wavelength$

$Ratio of Wavelengths$

$Substituting Values$

$Final Calculation$

$Conclusion$

The Sigma Insight: Matter Waves and de Broglie Relation

The Setup

The Master Equation: de Broglie Wavelength

The Ratio

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } de-Broglie Wavelength

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Wavelengths

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion