A proton and a Li3+ nucleus are accelerated by the same potential. If lambda_Li and lambda_p denote

Q: Why do we use $K = qV$ for kinetic energy?

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: Why is the charge of the Lithium nucleus $3e$?

The atomic number of Lithium is 3, meaning it has 3 protons in its nucleus. Since neutrons have no charge, the total charge of the nucleus is simply the charge of its 3 protons, which is $+3e$.

Q: Is it safe to approximate $24.9$ to $25$?

Yes! In competitive exams like JEE, when you encounter numbers like $24.9$ inside a square root, it is a deliberate hint from the examiner to approximate it to the nearest perfect square, $25$, to make the calculation clean and manageable.

Q: What happens if the particles were accelerated by different potentials?

If the potentials were different, say $V_1$ and $V_2$, they wouldn't cancel out in the ratio. You would have an extra factor of $\sqrt{V_2 / V_1}$ in your final expression.

Q: Why does a heavier particle have a shorter wavelength?

According to de-Broglie's relation $\lambda = \frac{h}{p}$, wavelength is inversely proportional to momentum. A heavier particle, when accelerated through the same potential, acquires more momentum, resulting in a shorter, more 'compressed' wave.

Question

A proton and a $	ext{Li}^{3+}$ nucleus are accelerated by the same potential. If $\lambda_{	ext{Li}}$ and $\lambda_p$ denote the de-Broglie wavelengths of $	ext{Li}^{3+}$ and proton respectively, then the value of $\frac{\lambda_{	ext{Li}}}{\lambda_p}$ is $x 	imes 10^{-1}$. The value of $x$ is ......... .
(Rounded off to the nearest integer)
(Mass of $	ext{Li}^{3+} = 8.3$ mass of proton)

Sigma Percentile · Accepted Answer

### The Setup: Particles in a Potential Field

Imagine a fascinating race at the subatomic level. On one track, we have a solitary proton ($p^+$). On the parallel track, a much heavier contender: a Lithium nucleus ($	ext{Li}^{3+}$). Both particles start from rest and are subjected to the exact same accelerating potential difference, $V$.

When a charged particle accelerates through a potential difference, the electrical work done on it is entirely converted into its kinetic energy. This fundamental principle of electrostatics gives us our first crucial equation:

$$K = qV$$

Here, $K$ is the kinetic energy, $q$ is the charge of the particle, and $V$ is the accelerating potential.

### The Master Equation: Bridging Particles and Waves

Now, we enter the quantum realm. According to Louis de Broglie's revolutionary hypothesis, every moving particle exhibits wave-like properties. The wavelength $\lambda$ of this matter wave is inversely proportional to the particle's momentum $p$:

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

Since momentum can be expressed in terms of kinetic energy as $p = \sqrt{2mK}$, we can substitute our earlier expression for kinetic energy ($K = qV$) into the de Broglie equation. This yields a powerful master equation that directly links the quantum wavelength to the macroscopic accelerating potential:

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

### The Ratio: Elegance in Cancellation

Our goal is to find the ratio of the wavelength of the Lithium nucleus to that of the proton, $\frac{\lambda_{	ext{Li}}}{\lambda_p}$. Let's set up the ratio using our master equation:

$$\frac{\lambda_{	ext{Li}}}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_{	ext{Li}}q_{	ext{Li}}V}}}{\frac{h}{\sqrt{2m_pq_pV}}}$$

Notice the mathematical elegance here! Planck's constant $h$, the number $2$, and the accelerating potential $V$ are identical for both particles. They beautifully cancel out, leaving us with a much simpler relationship depending only on the mass and charge of the particles:

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

### The Final Calculation: The Art of Approximation

Now, we plug in the specific values given in the problem. The charge of a proton is $e$, and the charge of the $	ext{Li}^{3+}$ nucleus is $3e$. The problem also explicitly states that the mass of the Lithium nucleus is $8.3$ times the mass of a proton ($m_{	ext{Li}} = 8.3 m_p$). Substituting these into our ratio:

$$\frac{\lambda_{	ext{Li}}}{\lambda_p} = \sqrt{\frac{m_p \cdot e}{8.3 m_p \cdot 3e}}$$

The mass of the proton ($m_p$) and the elementary charge ($e$) cancel out from the numerator and denominator. We are left with a pure numerical fraction:

$$\frac{\lambda_{	ext{Li}}}{\lambda_p} = \sqrt{\frac{1}{8.3 	imes 3}} = \sqrt{\frac{1}{24.9}}$$

Here is where a bit of exam intuition comes into play. The number $24.9$ is incredibly close to $25$, which is a perfect square. In competitive physics and chemistry, this is a deliberate signal to approximate:

$$\frac{\lambda_{	ext{Li}}}{\lambda_p} \approx \sqrt{\frac{1}{25}} = \frac{1}{5} = 0.2$$

The question asks for the answer in the format $x 	imes 10^{-1}$. Since $0.2$ is exactly $2 	imes 10^{-1}$, we can confidently conclude that **$x = 2$**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Kinetic Energy Gained$

$de-Broglie Wavelength$

$Master Equation$

$Ratio of Wavelengths$

$Substituting Values$

$Simplifying the Expression$

$Approximation$

$Final Answer$

The Sigma Insight: Wave Particle Duality

The Setup

The Master Equation

The Ratio

The Final Calculation

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Visualized Solution

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

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

.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; } Master Equation

.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; } Simplifying the Expression

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

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