An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will

Q: Why do we look for the 'shortest' wavelength?

Because wavelength is inversely proportional to energy ($E = \frac{hc}{\lambda}$). The 'shortest' wavelength corresponds to the 'maximum' energy transfer, which occurs when an electron loses 100% of its kinetic energy in a single collision.

Q: Can the emitted wavelength be longer than this calculated value?

Yes, absolutely. If the electron loses its energy gradually over multiple collisions, it will emit multiple photons, each with lower energy and therefore a longer wavelength. The value we calculated is the absolute minimum boundary.

Q: Why use 1240 instead of the actual values of $h$ and $c$?

It is a derived constant ($hc \approx 1240 \text{ eV}\cdot\text{nm}$) that drastically simplifies calculations. In competitive exams, saving time and avoiding arithmetic errors is crucial.

Q: What is the unit of the answer when using the 1240 shortcut?

When you use 1240, the resulting wavelength is strictly in nanometers (nm). If you prefer to work in Angstroms ($\text{\AA}$), you should use 12400 instead.

Q: Does the target material affect this shortest wavelength?

No. The continuous X-ray cut-off wavelength ($\lambda_{\text{min}}$) depends entirely on the accelerating voltage applied across the tube. The target material only affects the characteristic X-ray peaks, not the continuous spectrum boundary.

Question

An X-ray tube is operated at $1.24$ million volt. The shortest wavelength of the produced photon will be

Options:

(A) $10^{-3}	ext{ nm}$

(B) $10^{-1}	ext{ nm}$

(C) $10^{-2}	ext{ nm}$

(D) $10^{-4}	ext{ nm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine the inner workings of an X-ray tube. We have a cathode emitting electrons and a dense metal anode acting as a target. When we apply a massive potential difference of $1.24 	imes 10^6 	ext{ V}$ across this tube, the electrons are accelerated to incredibly high speeds.

The kinetic energy gained by an electron is directly proportional to the accelerating voltage, given by the equation $K_{	ext{max}} = eV$. In this case, the electron acquires a staggering kinetic energy of $1.24 	imes 10^6 	ext{ eV}$.

### The Master Equation
When this high-energy electron smashes into the target, it rapidly decelerates. According to the conservation of energy, this lost kinetic energy is converted into electromagnetic radiation—an X-ray photon.

To find the **shortest possible wavelength** (also known as the cut-off wavelength, $\lambda_{	ext{min}}$), we must assume the extreme case: the electron loses *all* of its kinetic energy in a single, catastrophic collision. This produces a single photon with the maximum possible energy.

Mathematically, this is expressed as:
$$E_{	ext{photon}} = \frac{hc}{\lambda_{	ext{min}}} = eV$$

### The Shortcut to Success
While you could plug in the standard SI values for Planck's constant ($h$) and the speed of light ($c$), doing so in a high-pressure exam like JEE is a recipe for calculation errors. Instead, we use a powerful derived shortcut:

$$\lambda_{	ext{min}} (	ext{in nm}) = \frac{1240 	ext{ eV nm}}{E (	ext{in eV})}$$

Since our energy $E$ is exactly $1.24 	imes 10^6 	ext{ eV}$, we can substitute this directly into our shortcut formula.

### Final Calculation
Let's perform the substitution:

$$\lambda_{	ext{min}} = \frac{1240}{1.24 	imes 10^6} 	ext{ nm}$$

Notice how perfectly the numbers are designed to cancel out. We can rewrite the denominator to match the numerator:

$$\lambda_{	ext{min}} = \frac{1240}{1240 	imes 10^3} 	ext{ nm}$$

The $1240$ cancels out beautifully, leaving us with:

$$\lambda_{	ext{min}} = \frac{1}{10^3} 	ext{ nm} = 10^{-3} 	ext{ nm}$$

This elegant cancellation is a hallmark of well-designed JEE problems. By mastering the $1240$ shortcut, you transform a tedious calculation into a ten-second mental math exercise.

Animated Solution for Physics - Dual Nature of Matter and Radiation: An X-ray tube is operated at $1.24$ million volt. The shortest wavelength of the produced photon will be

Select Answer:

Visualized Solution

$The X-ray Tube Setup$

$Shortest Wavelength (Cut-off Wavelength)$

$The Shortcut Formula$

$Substitution and Calculation$

$Final Answer$

$The Way Forward$

The Sigma Insight: Photon Theory of Light

Analyzing the Setup Imagine the inner workings of an X-ray tube

The Master Equation When this high-energy electron smashes into the target, it rapidly decelerates

The Shortcut to Success While you could plug in the standard SI values for Planck's constant ( $h$ ) and the speed of light ( $c$ ), doing so in a high-pressure exam like JEE is a recipe for calculation errors

Final Calculation

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Animated Solution for Physics - Dual Nature of Matter and Radiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be