In an X-ray tube, electrons emitted from a filament (cathode) carrying current I hit a target (anode

Q: Why doesn't the distance $d$ affect the cut-off wavelength?

The cut-off wavelength depends on the maximum kinetic energy of the electrons when they hit the target. This energy is equal to the work done by the electric field, which is $W = qV$. Notice that distance $d$ is not in this equation! Changing $d$ alters the electric field strength ($E = V/d$) and the time it takes to cross the gap, but the total energy gained remains exactly $eV$.

Q: What is the difference between the filament current and the accelerating voltage?

Great question! The filament current heats up the cathode, causing it to boil off electrons (thermionic emission). It controls the *quantity* of electrons. The accelerating voltage, on the other hand, pulls these electrons towards the anode. It controls the *energy* of each individual electron.

Q: Why do the characteristic peaks stay at the exact same wavelengths?

Characteristic X-rays are produced when an incoming electron knocks out an inner-shell electron of the target atom, and an outer-shell electron falls in to fill the gap. The energy difference between these specific atomic shells is a fixed property of the target material (like a fingerprint). Since we didn't change the target metal, these energy gaps—and thus the wavelengths—remain identical.

Q: Does the intensity of characteristic X-rays decrease too?

Yes, absolutely! While their wavelengths don't change, their intensity does. Intensity is simply the number of photons emitted per second. Since we halved the filament current, fewer electrons are striking the target, meaning fewer inner-shell collisions occur. Thus, the peaks will be shorter in height.

Q: What would happen if we changed the target material to one with a higher atomic number?

According to Moseley's Law ($\sqrt{\nu} \propto Z - b$), a higher atomic number $Z$ means the energy gaps between shells are larger. This results in emitted X-ray photons with higher frequency and shorter wavelengths. So, the characteristic peaks would shift to the left (towards shorter wavelengths) on the graph!

Question

In an X-ray tube, electrons emitted from a filament (cathode) carrying current $I$ hit a target (anode) at a distance $d$ from the cathode. The target is kept at a potential $V$ higher than the cathode resulting in emission of continuous and characteristic X-rays. If the filament current $I$ is decreased to $\frac{I}{2}$, the potential difference $V$ is increased to $2V$, and the separation distance $d$ is reduced to $\frac{d}{2}$, then

Options:

(A) the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same

(B) the cut-off wavelength as well as the wavelengths of the characteristic X-rays will remain the same

(C) the cut-off wavelength will reduce to half, and the intensities of all the X-rays will decrease

(D) the cut-off wavelength will become two times larger, and the intensity of all the X-rays will decrease

Sigma Percentile · Accepted Answer

### Unveiling the Secrets of the X-Ray Spectrum

Imagine you are standing in front of an X-ray tube, ready to tweak its controls. The problem presents us with a scenario where we manipulate three distinct parameters: the accelerating voltage $V$ is doubled, the filament current $I$ is halved, and the distance $d$ between the cathode and anode is halved. Our mission is to decode exactly how these changes sculpt the resulting X-ray spectrum.

### The Continuous Spectrum and the Cut-off Wavelength

Let's first tackle the continuous part of the spectrum, often called Bremsstrahlung (braking radiation). When electrons are accelerated across the tube, they gain a maximum kinetic energy equal to the work done by the electric field, which is $eV$. If an electron loses all this energy in a single collision, it emits an X-ray photon with the maximum possible energy, and consequently, the minimum possible wavelength.

This is governed by the master equation:

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

Notice the inverse relationship between the cut-off wavelength $\lambda_{\min}$ and the accelerating voltage $V$. By doubling the voltage to $2V$, we are pumping twice as much energy into the electrons. As a direct mathematical consequence, the new cut-off wavelength becomes exactly half of its original value:

$$\lambda_{\min}' = \frac{hc}{e(2V)} = \frac{\lambda_{\min}}{2}$$

This means the entire X-ray spectrum shifts to the left, starting at a much shorter wavelength.

### The Unshakable Characteristic X-Rays

Superimposed on the continuous curve are sharp, intense spikes known as characteristic X-rays. These are born from a completely different mechanism: an incoming electron knocks out an inner-shell electron of the target atom, and an outer-shell electron cascades down to fill the void, releasing a photon.

The energy of this photon—and therefore its wavelength—is dictated strictly by the atomic energy levels of the target material. According to Moseley's Law, this is a fundamental property of the element's atomic number $Z$. Because we did not swap out the target metal in our experiment, these energy gaps remain identical. Therefore, **the wavelengths of the characteristic X-rays remain absolutely unchanged.**

### The Flow of Electrons and Intensity

Now, let's address the overall intensity of the X-rays. Intensity is a measure of how many X-ray photons are being emitted per second. This is directly proportional to the number of electrons striking the target per second.

Here is where the filament current $I$ plays its crucial role. The filament current heats the cathode, controlling the rate of thermionic emission. By halving the filament current to $I/2$, we drastically reduce the number of electrons boiling off the cathode. Fewer striking electrons mean fewer X-ray photons are produced. Consequently, **the intensity of all X-rays—both continuous and characteristic—will decrease.**

### The Distractor: Distance $d$

What about the distance $d$ being halved? This is a classic trap! While bringing the anode closer increases the electric field strength ($E = V/d$), the total work done on an electron crossing the gap is still Force $	imes$ Distance, which evaluates to $(eV/d) 	imes d = eV$. The final kinetic energy of the electrons upon impact is completely unaffected by $d$. It is purely a distractor.

By synthesizing these physical truths, we can confidently conclude that the cut-off wavelength halves, the characteristic wavelengths stay put, and the overall intensity drops. This perfectly aligns with options (A) and (C).

Select Answer:

Visualized Solution

\text{X-Ray Tube Parameters}

\text{Cut-off Wavelength } (\lambda_{\min})

\text{Effect of Doubling Voltage}

\text{Characteristic Wavelengths}

\text{Intensity of X-rays}

\text{Effect of Halving Current}

\text{Conclusion}

The Sigma Insight: Photon Theory of Light

Unveiling the Secrets of the X-Ray Spectrum

The Continuous Spectrum and the Cut-off Wavelength

The Unshakable Characteristic X-Rays

The Flow of Electrons and Intensity

The Distractor

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