A particle moving with kinetic energy E has de Broglie wavelength lambda. If energy Delta E is added

Q: Why do we use the formula with kinetic energy instead of just momentum?

The problem explicitly gives us the initial state in terms of kinetic energy $E$ and asks for the added energy $\Delta E$. While $\lambda = \frac{h}{p}$ is the fundamental equation, substituting $p = \sqrt{2mK}$ allows us to directly relate the given variables without needing to calculate momentum as an intermediate step.

Q: Does the mass of the particle change when we add energy?

No, in non-relativistic classical mechanics (which is assumed here unless stated otherwise), the rest mass $m$ of the particle remains constant regardless of its kinetic energy. Only its velocity, and consequently its momentum and kinetic energy, change.

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

According to the de Broglie relation $\lambda = \frac{h}{p}$, wavelength is inversely proportional to momentum. Higher kinetic energy means higher velocity and higher momentum. As the denominator (momentum) grows larger, the overall fraction (wavelength) becomes smaller.

Q: Can I just say that since wavelength is halved, energy must be doubled?

No, that's a very common trap! Wavelength is inversely proportional to the *square root* of kinetic energy, not the kinetic energy itself. To halve the wavelength, the kinetic energy must increase by a factor of $2^2$, which is $4$. So the final energy is $4E$, meaning you added $3E$.

Q: Would this calculation be different if the particle was a photon?

Yes, absolutely! For a photon, energy is directly related to wavelength by $E = \frac{hc}{\lambda}$. If you halve the wavelength of a photon, its energy exactly doubles (becomes $2E$). But for a particle with mass, the relationship involves a square root, leading to a different result.

Question

A particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy, the wavelength become $\lambda /2$. Value of $\Delta E$ is

Options:

(A) $2E$

(B) $4E$

(C) $3E$

(D) $E$

Sigma Percentile · Accepted Answer

### The Quantum Dance of Energy and Wavelength

Imagine a tiny particle zooming through space. In the classical world, it's just a chunk of mass moving with some velocity. But in the quantum realm, this moving particle behaves like a wave. This beautiful duality is captured by the **de Broglie wavelength**.

The problem presents us with a particle that has an initial kinetic energy $E$ and a corresponding de Broglie wavelength $\lambda$. We are then asked a fascinating question: If we want to compress this wave, specifically to halve its wavelength to $\lambda/2$, how much extra energy $\Delta E$ do we need to pump into the particle?

### The Master Equation

To solve this, we need the mathematical bridge between the wave world and the particle world. The de Broglie wavelength $\lambda$ is given by Planck's constant $h$ divided by the particle's momentum $p$:

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

However, our problem speaks in terms of kinetic energy $K$, not momentum. We know that kinetic energy $K = \frac{p^2}{2m}$, which means momentum $p = \sqrt{2mK}$. Substituting this into our de Broglie equation gives us our master tool:

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

### Setting Up the Scenarios

Let's translate the problem's two states into mathematics.

**State 1 (Initial):** The particle has kinetic energy $E$. Its wavelength is:
$$\lambda = \frac{h}{\sqrt{2mE}}$$

**State 2 (Final):** We add an energy $\Delta E$, making the total kinetic energy $E + \Delta E$. The new wavelength is $\lambda/2$:
$$\frac{\lambda}{2} = \frac{h}{\sqrt{2m(E + \Delta E)}}$$

### The Elegant Execution

We have a system of two equations. The most elegant way to solve for $\Delta E$ is to divide the first equation by the second. This brilliant move instantly annihilates the constants $h$ and $2m$, which we don't know and don't need to know.

$$\frac{\lambda}{\frac{\lambda}{2}} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{h}{\sqrt{2m(E + \Delta E)}}}$$

The left side simplifies to $2$. On the right side, the fractions flip and multiply, leaving us with a single square root:

$$2 = \sqrt{\frac{E + \Delta E}{E}}$$

To liberate our variables from the square root, we square both sides:

$$4 = \frac{E + \Delta E}{E}$$

Now, it's a straightforward algebraic sprint to the finish line. Multiply both sides by $E$:

$$4E = E + \Delta E$$

Subtract $E$ from $4E$, and we arrive at our final answer:

$$\Delta E = 3E$$

### The Intuitive Takeaway

Let's step back and look at what this means. The formula $\lambda = \frac{h}{\sqrt{2mK}}$ tells us that wavelength is inversely proportional to the square root of kinetic energy ($\lambda \propto \frac{1}{\sqrt{K}}$).

If you want to divide the wavelength by $2$, you must multiply the denominator (the square root) by $2$. To make a square root twice as large, the value *inside* the square root must become $4$ times larger.

Therefore, the final kinetic energy must be $4E$. Since the particle already had an energy of $E$, the *extra* energy you must add is simply $4E - E = 3E$. This intuitive check confirms our mathematical derivation perfectly!

Animated Solution for Physics - Dual Nature of Matter and Radiation: A particle moving with kinetic energy $E$ has de Broglie wavelength $λ$ . If energy $Δ E$ is added to its energy, the wavelength become $λ /2$ . Value of $Δ E$ is

Select Answer:

Visualized Solution

Initial State

de Broglie Wavelength Formula

Setting up the Equations

Dividing the Equations

Squaring Both Sides

Solving for $Δ E$

Conclusion

The Sigma Insight: Matter Waves and de Broglie Relation

The Quantum Dance of Energy and Wavelength

The Master Equation

Setting Up the Scenarios

The Elegant Execution

The Intuitive Takeaway

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