An electron having de-Broglie wavelength lambda is incident on a target in a X-ray tube. Cut-off wav

Q: Why do we use the de-Broglie wavelength for the electron?

Because the electron is a quantum particle, and its momentum is intrinsically linked to its matter wave (de-Broglie wavelength) via the relation $p = h/\lambda$.

Q: What exactly is the 'cut-off' wavelength?

It is the absolute minimum wavelength an emitted X-ray can have. This happens when the electron loses 100% of its kinetic energy in a single collision, converting it entirely into one highly energetic photon.

Q: Why did we use $K = p^2/2m$ instead of $1/2 mv^2$?

While both are correct, we are given the de-Broglie wavelength $\lambda$, which directly gives us momentum $p$. Using $p^2/2m$ is a much faster and more elegant mathematical shortcut.

Q: Can the emitted X-ray have a wavelength longer than $\lambda_c$?

Yes! If the electron only loses a fraction of its energy in a collision, the resulting photon will have less energy, which corresponds to a longer wavelength. $\lambda_c$ is just the extreme lower limit.

Q: Does the mass of the target matter in this equation?

For the continuous X-ray spectrum (which gives the cut-off wavelength), the target material doesn't matter. The cut-off depends solely on the initial kinetic energy of the incident electrons.

Question

An electron having de-Broglie wavelength $\lambda$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is

Options:

(A) Zero

(B) $\frac{2m^2c^2\lambda^2}{h^2}$

(C) $\frac{2mc\lambda^2}{h}$

(D) $\frac{hc}{mc}$

Sigma Percentile · Accepted Answer

### The Setup: A Quantum Bullet
Imagine an electron zooming towards a metal target inside an X-ray tube. In the quantum world, this electron isn't just a tiny particle; it behaves like a wave. This wave nature is described by its de-Broglie wavelength, $\lambda$. When this high-speed electron smashes into the heavy metal target, a dramatic energy conversion takes place. The electron rapidly decelerates, losing its kinetic energy. According to the conservation of energy, this lost kinetic energy doesn't just vanish—it is radiated away as an electromagnetic wave, specifically an X-ray photon.

### The Energy Translation
To find out the properties of this emitted X-ray, we first need to understand exactly how much energy the electron is carrying. We start with the fundamental de-Broglie relation, which connects the electron's wavelength to its momentum:

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

Now, we need to express the electron's kinetic energy ($K$) in terms of this momentum. The classic formula $K = \frac{1}{2}mv^2$ is great, but the momentum-based version is much more elegant here:

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

Substituting our quantum momentum into this equation, we get the total kinetic energy of the incident electron:

$$K = \frac{\left(\frac{h}{\lambda}ight)^2}{2m} = \frac{h^2}{2m\lambda^2}$$

### The Birth of an X-Ray
When the electron hits the target, it can lose its energy in multiple small collisions, emitting several lower-energy photons. However, we are looking for the **cut-off wavelength** ($\lambda_c$). This represents the absolute minimum wavelength, which corresponds to the absolute maximum energy. This extreme case happens only when the electron loses 100% of its kinetic energy in a single, catastrophic collision, transferring all of it to a single X-ray photon.

The energy of this photon is given by Planck's equation:

$$E_{	ext{photon}} = \frac{hc}{\lambda_c}$$

### The Grand Equation
By equating the maximum kinetic energy of the electron to the energy of the emitted photon, we set up our master equation:

$$K = E_{	ext{photon}}$$

$$\frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_c}$$

Notice the beautiful symmetry here. We can cancel one Planck's constant ($h$) from both sides:

$$\frac{h}{2m\lambda^2} = \frac{c}{\lambda_c}$$

Finally, we rearrange the terms to isolate the cut-off wavelength $\lambda_c$:

$$\lambda_c = \frac{2mc\lambda^2}{h}$$

And there we have it! The minimum possible wavelength of the emitted X-ray is directly proportional to the square of the electron's de-Broglie wavelength. This elegant derivation perfectly bridges matter waves and electromagnetic radiation.

Animated Solution for Physics - Dual Nature of Matter and Radiation: An electron having de-Broglie wavelength $λ$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is

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Visualized Solution

$The Physical Setup$

$Electron Momentum$

$Electron Kinetic Energy$

$X-ray Cut-off Wavelength$

$Equating Energies$

$Final Answer$

The Sigma Insight: Matter Waves and de Broglie Relation

The Setup

The Energy Translation To find out the properties of this emitted X-ray, we first need to understand exactly how much energy the electron is carrying

The Birth of an X-Ray When the electron hits the target, it can lose its energy in multiple small collisions, emitting several lower-energy photons

The Grand Equation

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X-ray Cut-off Wavelength

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

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