If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor
(a

Q: Why does the wavelength decrease when the kinetic energy increases?

Kinetic energy is a measure of motion. Higher kinetic energy means the electron is moving faster, which means it has a higher momentum. According to de-Broglie's relation ($\lambda = \frac{h}{p}$), wavelength is inversely proportional to momentum. Therefore, a higher momentum results in a shorter, more compressed wavelength.

Q: Can I use the formula $E = mc^2$ to find the energy here?

No, $E = mc^2$ represents the rest mass energy (or total relativistic energy). For a free electron moving at non-relativistic speeds (much less than the speed of light), we use the classical kinetic energy formula $K = \frac{1}{2}mv^2$ or $K = \frac{p^2}{2m}$.

Q: What if the question said the velocity was doubled instead of the kinetic energy?

If the velocity $v$ is doubled, the momentum $p = mv$ is also exactly doubled. Since $\lambda = \frac{h}{p}$, doubling the momentum would make the wavelength exactly half ($\frac{\lambda}{2}$). However, doubling the kinetic energy only increases the momentum by a factor of $\sqrt{2}$.

Q: Does the mass of the electron change when its kinetic energy increases?

In classical mechanics and for speeds much lower than the speed of light, the mass of the electron is considered constant. Relativistic mass increase only becomes significant when the particle's speed approaches the speed of light.

Q: Is this de-Broglie formula applicable to macroscopic objects like a moving car?

Yes, the formula applies to all matter. However, because Planck's constant ($h$) is incredibly small ($\approx 6.63 \times 10^{-34}$ J s) and a car has a massive momentum compared to an electron, the resulting wavelength is so infinitesimally small that it is completely unobservable and has no practical effect on the car's macroscopic behavior.

Question

If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor

Options:

(A) $\frac{1}{2}$

(B) $2$

(C) $\frac{1}{\sqrt{2}}$

(D) $\sqrt{2}$

Sigma Percentile · Accepted Answer

### The Dance of the Electron: Kinetic Energy and de-Broglie Wavelength

Imagine an electron not just as a tiny, solid billiard ball, but as a dynamic, oscillating wave traveling through space. This is the beautiful reality of quantum mechanics introduced by Louis de Broglie. Every moving particle has a wave associated with it, and the properties of this wave are intimately tied to the particle's motion.

In this problem, we are asked to find out how the de-Broglie wavelength of a free electron changes when its kinetic energy is doubled. To solve this, we need to build a bridge between the particle world (kinetic energy) and the wave world (wavelength).

### The Master Equation

The fundamental connection between these two worlds is the de-Broglie equation:

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

Here, $\lambda$ is the wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle.

However, our question gives us information about **kinetic energy ($K$)**, not momentum. So, we need to express momentum in terms of kinetic energy. Recall the standard formula for kinetic energy:

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

If we multiply and divide the right side by the mass $m$, we get:

$$K = \frac{m^2v^2}{2m} = \frac{p^2}{2m}$$

Rearranging this to solve for momentum $p$, we find:

$$p = \sqrt{2mK}$$

Now, let's substitute this expression for momentum back into our de-Broglie equation. This gives us our master equation for the problem:

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

### Analyzing the Change

Look closely at the master equation. Planck's constant ($h$) and the mass of the electron ($m$) are constants. Therefore, the wavelength is inversely proportional to the square root of the kinetic energy:

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

This proportionality is the key. It tells us exactly how $\lambda$ responds to changes in $K$. Let's set up a ratio for the final state (state 2) compared to the initial state (state 1):

$$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}}$$

The problem states that the kinetic energy is doubled, meaning **$K_2 = 2K_1$**. Let's substitute this into our ratio:

$$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{2K_1}}$$

The $K_1$ terms cancel out beautifully, leaving us with:

$$\frac{\lambda_2}{\lambda_1} = \frac{1}{\sqrt{2}}$$

### Conclusion

This result means that the new de-Broglie wavelength $\lambda_2$ is $\frac{1}{\sqrt{2}}$ times the original wavelength $\lambda_1$. As the electron gains energy and speeds up, its associated matter wave gets compressed, resulting in a shorter wavelength.

Understanding this inverse square root relationship is crucial for quickly solving many problems in modern physics!

Animated Solution for Physics - Dual Nature of Matter and Radiation: If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor

Select Answer:

Visualized Solution

Visualizing the Electron Wave

de-Broglie Wavelength

Momentum and Kinetic Energy

The Master Equation

Applying the Condition

Calculating the Ratio

Final Answer

The Way Forward

The Sigma Insight: Matter Waves and de Broglie Relation

The Dance of the Electron

The Master Equation

Analyzing the Change

Conclusion

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Animated Solution for Physics - Dual Nature of Matter and Radiation: If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor

Select Answer:

Visualized Solution

Visualizing the Electron Wave

de-Broglie Wavelength

Momentum and Kinetic Energy

The Master Equation

Applying the Condition

Calculating the Ratio

Final Answer

The Way Forward

The Sigma Insight: Matter Waves and de Broglie Relation

The Dance of the Electron

The Master Equation

Analyzing the Change

Conclusion

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