A potential difference of 20 kV is applied across an X-ray tube. The minimum wavelength of X-rays ge

Q: Why is there a minimum wavelength in the X-ray spectrum?

The minimum wavelength corresponds to the maximum energy a photon can have. This happens when an electron transfers 100% of its kinetic energy into a single photon during a collision. Since the electron's kinetic energy is finite (determined by the accelerating voltage), the photon's energy is capped, meaning its wavelength cannot be shorter than a certain limit.

Q: Should I use 12400 or 12375 for the value of hc?

Both are acceptable approximations. 12400 is easier for mental math, while 12375 is slightly more precise. In this specific problem, using 12375 gives 0.61875 (rounds to 0.62), and using 12400 gives 0.62. Both lead to the same rounded answer.

Q: Does the minimum wavelength depend on the target material?

No, the minimum wavelength (cut-off wavelength) depends ONLY on the accelerating potential difference $V$. It is completely independent of the target material. However, the characteristic X-ray peaks do depend on the target material.

Q: What happens if we increase the potential difference?

If you increase the potential difference $V$, the electrons gain more kinetic energy. Consequently, the maximum photon energy increases, which means the minimum wavelength $\lambda_{\text{min}}$ will decrease (shift to the left on the spectrum graph).

Q: Why do we convert kV to V before substituting?

The shortcut formula $\lambda_{\text{min}} = \frac{12375}{V}$ specifically requires the potential $V$ to be in standard Volts (V) to output the wavelength in Angstroms ($\text{\AA}$). If you leave it in kV, your answer will be off by a factor of 1000.

Question

A potential difference of $20	ext{ kV}$ is applied across an X-ray tube. The minimum wavelength of X-rays generated is ....... $	ext{\AA}$.

Sigma Percentile · Accepted Answer

### The Physics of X-ray Production

Imagine you are inside an X-ray tube. Electrons are boiled off a cathode and accelerated across a massive potential difference towards a metal target. As they slam into the target, they decelerate violently. According to classical electromagnetism, an accelerating (or decelerating) charge emits electromagnetic radiation. This is the origin of the continuous X-ray spectrum, also known as **Bremsstrahlung** or 'braking radiation'.

### The Cut-off Wavelength

When an electron hits the target, it can lose its kinetic energy in multiple collisions, emitting several lower-energy photons. However, the most extreme case occurs when an electron loses **all** of its kinetic energy in a single, head-on collision with a target atom.

In this scenario, a single photon is emitted carrying the maximum possible energy. Since the energy of a photon is inversely proportional to its wavelength ($E = \frac{hc}{\lambda}$), this maximum energy corresponds to the **minimum possible wavelength**, often called the cut-off wavelength ($\lambda_{	ext{min}}$).

### The Master Equation

The kinetic energy gained by the electron is simply the charge of the electron multiplied by the accelerating potential: $K = eV$. Equating this to the maximum photon energy gives us the Duane-Hunt law:

$$eV = \frac{hc}{\lambda_{	ext{min}}}$$

Rearranging for $\lambda_{	ext{min}}$:

$$\lambda_{	ext{min}} = \frac{hc}{eV}$$

### The Calculation Shortcut

In competitive exams like JEE, speed is crucial. Instead of plugging in the standard SI values for Planck's constant ($h$) and the speed of light ($c$), we use a highly convenient shortcut. The product $hc$ is approximately $12400 	ext{ eV}\cdot	ext{\AA}$ (or more precisely, $12375 	ext{ eV}\cdot	ext{\AA}$).

This transforms our equation into a beautifully simple form:

$$\lambda_{	ext{min}} (	ext{in \AA}) = \frac{12375}{V (	ext{in volts})}$$

### Final Execution

For our specific problem, the accelerating potential is $20 	ext{ kV}$, which is $20,000 	ext{ V}$. Substituting this into our shortcut formula:

$$\lambda_{	ext{min}} = \frac{12375}{20000} = 0.61875 	ext{ \AA}$$

Rounding to two decimal places, we get our final answer:

**$\lambda_{	ext{min}} \approx 0.62 	ext{ \AA}$**

This elegant result tells us the absolute shortest wavelength of X-rays that this specific tube can produce under a $20 	ext{ kV}$ potential.

Animated Solution for Physics - Dual Nature of Matter and Radiation: A potential difference of $20 kV$ is applied across an X-ray tube. The minimum wavelength of X-rays generated is ....... $\overset{˚}{A}$ .

Enter Numerical Value:

Visualized Solution

$Energy Conservation in X-ray Production$

$The Duane-Hunt Formula$

$Substitution$

$Final Calculation$

The Sigma Insight: Photon Theory of Light

The Physics of X-ray Production

The Cut-off Wavelength

The Master Equation

The Calculation Shortcut

Final Execution

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