A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the p

Q: Why do we use the ratio instead of calculating individual wavelengths?

We don't have the exact velocities of the particles, only their relative relation ($v_p = 5v_e$). By taking a ratio, the unknown velocity $v_e$ cancels out completely, allowing us to solve the problem without needing its exact value.

Q: What is the mass of an electron? Is it always given?

The mass of an electron is a fundamental physical constant, $m_e = 9.1 \times 10^{-31}$ kg. While sometimes provided in the exam paper's data sheet, it is highly recommended to memorize it for JEE.

Q: Why does the particle have a much smaller wavelength than the electron?

Wavelength is inversely proportional to momentum ($p = mv$). Because the particle is moving 5 times faster and has a significantly larger mass than the electron, its momentum is much larger, resulting in a much smaller de-Broglie wavelength.

Q: Can we use the kinetic energy formula for de-Broglie wavelength here?

You could use $\lambda = \frac{h}{\sqrt{2mK}}$ if you knew the kinetic energies. However, since the problem explicitly gives the relationship between their velocities, using the momentum form $p = mv$ is much more direct and efficient.

Q: How can I handle this calculation quickly without a calculator?

Look for approximations. In the denominator, $5 \times 1.878$ is roughly $5 \times 1.9 = 9.5$ (exactly $9.39$). Then you have $9.1 / 9.39$, which is slightly less than $1$. This perfectly aligns with $0.97 \times 10^{-27}$, or $9.7 \times 10^{-28}$, matching option (d) without needing tedious long division.

Question

A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is $1.878 	imes 10^{-4}$. The mass of the particle is close to

Options:

(A) $4.8 	imes 10^{-27}$ kg

(B) $9.1 	imes 10^{-31}$ kg

(C) $1.2 	imes 10^{-28}$ kg

(D) $9.7 	imes 10^{-28}$ kg

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine we have two distinct entities in motion: our well-known electron and an unknown particle. The problem states a clear kinematic relationship between them. The unknown particle is moving significantly faster—exactly five times the speed of the electron.

Mathematically, we can express this as:
$$v_p = 5v_e$$

Whenever a particle is in motion, quantum mechanics tells us that it exhibits wave-like properties. This is governed by the de-Broglie wavelength, which inversely relates the wavelength to the particle's momentum.

### The Master Equation
The fundamental tool we need here is the de-Broglie wavelength formula:
$$\lambda = \frac{h}{p} = \frac{h}{mv}$$

We are provided with the ratio of the de-Broglie wavelength of the particle to that of the electron. Let's set up this ratio carefully. By dividing the wavelength of the particle by the wavelength of the electron, the Planck's constant ($h$) gracefully cancels out:
$$\frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{m_p v_p}}{\frac{h}{m_e v_e}} = \frac{m_e v_e}{m_p v_p}$$

### Raw Setup and Substitution
Now, we substitute the given values into our ratio equation. We know the ratio is $1.878 	imes 10^{-4}$, and we can replace $v_p$ with $5v_e$:
$$1.878 	imes 10^{-4} = \frac{m_e v_e}{m_p (5v_e)}$$

Notice how the velocity of the electron ($v_e$) appears in both the numerator and the denominator. This is a beautiful moment in physics problems where an unknown variable simply vanishes, leaving us with a clean relationship between the masses:
$$1.878 	imes 10^{-4} = \frac{m_e}{5 m_p}$$

### Final Calculation
Our goal is to find the mass of the unknown particle, $m_p$. Let's rearrange the equation to isolate $m_p$:
$$m_p = \frac{m_e}{5 	imes 1.878 	imes 10^{-4}}$$

To proceed, we must recall the standard mass of an electron, which is $m_e = 9.1 	imes 10^{-31} 	ext{ kg}$. Substituting this value in, we get:
$$m_p = \frac{9.1 	imes 10^{-31}}{5 	imes 1.878 	imes 10^{-4}}$$

Let's simplify the denominator first. Multiplying $5$ by $1.878$ gives exactly $9.39$. Now the expression looks much more manageable:
$$m_p = \frac{9.1 	imes 10^{-31}}{9.39 	imes 10^{-4}}$$

Dividing $9.1$ by $9.39$ yields approximately $0.969$. Adjusting the powers of $10$ ($-31 - (-4) = -27$), we get:
$$m_p \approx 0.969 	imes 10^{-27} 	ext{ kg}$$

To match the standard scientific notation of our options, we shift the decimal point:
$$m_p \approx 9.7 	imes 10^{-28} 	ext{ kg}$$

**This perfectly matches option (d).**

Animated Solution for Physics - Dual Nature of Matter and Radiation: A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is $1.878 \times 1 0^{- 4}$ . The mass of the particle is close to

Select Answer:

Visualized Solution

Visualizing the Setup

de-Broglie Wavelength Formula

Setting up the Ratio

Substituting Known Values

Isolating the Unknown Mass

Final Calculation

Conclusion

The Sigma Insight: Matter Waves and de Broglie Relation

Analyzing the Setup Imagine we have two distinct entities in motion: our well-known electron and an unknown particle

The Master Equation

Raw Setup and Substitution Now, we substitute the given values into our ratio equation

Final Calculation Our goal is to find the mass of the unknown particle, $m_{p}$

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