The de-Broglie wavelength of a proton and alpha-particle are equal. The ratio of their velocities is

Q: Why is the mass of an alpha particle 4 times that of a proton?

An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons. Since protons and neutrons have approximately the same mass, the total mass is roughly 4 times the mass of a single proton.

Q: Does the charge of the particles matter in this question?

No, the de-Broglie wavelength formula $\lambda = \frac{h}{mv}$ only depends on mass and velocity, not charge. Charge would only matter if they were accelerated by a potential difference.

Q: What if their kinetic energies were equal instead of wavelengths?

If kinetic energies were equal, we would use the relation $\lambda = \frac{h}{\sqrt{2mK}}$. The ratio of wavelengths would then be inversely proportional to the square root of their masses.

Q: Can we use this formula for macroscopic objects?

Yes, the de-Broglie equation applies to all matter. However, for macroscopic objects, the mass is so large that the wavelength becomes undetectably small.

Q: Why did we cancel Planck's constant?

Planck's constant $h$ is a universal constant. Since it appears in the numerator on both sides of the equation $\frac{h}{p_p} = \frac{h}{p_\alpha}$, it can be mathematically canceled out.

Question

The de-Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is

Options:

(A) 4 : 3

(B) 4 : 1

(C) 4 : 2

(D) 1 : 4

Sigma Percentile · Accepted Answer

### Visualizing the Particles

Imagine we have two particles, a proton and an alpha particle. Focus on the setup. Let us assume the mass of the proton is $m_p$ and its velocity is $v_p$.

Similarly, let the mass of the alpha particle be $m_\alpha$ and its velocity be $v_\alpha$.

Visualizing these two particles helps us ground the abstract math into a physical reality. We are comparing a relatively light proton with a heavier alpha particle.

### The Master Equation: de-Broglie Wavelength

Now, recall the fundamental formula for de-Broglie wavelength. The wavelength $\lambda$ of any moving particle is equal to the ratio of Planck's constant $h$ to its momentum $p$.

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

This beautiful equation bridges the gap between the wave nature and particle nature of matter. It tells us that every moving object has an associated wave!

### Equating the Wavelengths

The question explicitly states that their de-Broglie wavelengths are exactly the same. So, let us equate their $\lambda$ values.

$$ \frac{h}{m_p v_p} = \frac{h}{m_\alpha v_\alpha} $$

Look closely at this equation. The Planck's constant $h$ is present in the numerator on both sides.

This means $h$ will simply cancel out, leaving us with a direct relationship between their momenta.

$$ m_p v_p = m_\alpha v_\alpha $$

This implies that for their wavelengths to be equal, their linear momenta must be perfectly identical!

### The Crucial Catch: Mass Relationship

There is a catch here, and it is a favorite concept for JEE. What exactly is an alpha particle?

An alpha particle is actually a helium nucleus, which consists of two protons and two neutrons. Since protons and neutrons have roughly the same mass, the total mass of an alpha particle is four times the mass of a single proton.

$$ m_\alpha = 4m_p $$

Don't make a silly mistake by ignoring this mass relationship. It is the key to unlocking the final answer.

### Final Calculation

Now, let us substitute this mass value back into our momentum equation.

$$ m_p v_p = 4m_p v_\alpha $$

Notice how $m_p$ appears on both sides? We can easily cancel it out.

$$ v_p = 4v_\alpha $$

Finally, rearranging the terms gives us the ratio of their velocities.

$$ \frac{v_p}{v_\alpha} = \frac{4}{1} $$

So, the ratio of the velocity of the proton to the velocity of the alpha particle is $4 : 1$. It is a very simple and elegant result!

Animated Solution for Physics - Dual Nature of Matter and Radiation: The de-Broglie wavelength of a proton and $α$ -particle are equal. The ratio of their velocities is

Select Answer:

Visualized Solution

$V i s u a l i z in g t h e P a r t i c l es$

$d e - B r o g l i e W a v e l e n g t h$

$E q u a t in g W a v e l e n g t h s$

$M o m e n t u m E q u a l i t y$

$M a ss R e l a t i o n s hi p$

$S u b s t i t u t in g M a ss$

$R a t i o o f V e l oc i t i es$

The Sigma Insight: Matter Waves and de Broglie Relation

Visualizing the Particles

The Master Equation: de-Broglie Wavelength

Equating the Wavelengths

The Crucial Catch

Final Calculation

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Animated Solution for Physics - Dual Nature of Matter and Radiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The de-Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is

Select Answer:

Visualized Solution

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

.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; } Equating Wavelengths

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

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

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

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