Q.Find the

Concept: De Broglie Wavelength — but here it’s the inverse: X-rays are produced when fast electrons are suddenly stopped. The maximum photon energy equals the kinetic energy of the electron.

1. The kinetic energy of an electron accelerated through $30\ \text{kV}$ is

   $K = eV = 30\ \text{keV}$.

2. The maximum X-ray photon energy is the same:

   $E_{\text{max}} = h f_{\text{max}} = eV$.

3. (a) Maximum frequency:

   $f_{\text{max}} = \frac{eV}{h} = \frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.63 \times 10^{-34}}$

   $= 7.24 \times 10^{18}\ \text{Hz}$.

4. (b) Minimum wavelength:

   $\lambda_{\text{min}} = \frac{c}{f_{\text{max}}} = \frac{3 \times 10^8}{7.24 \times 10^{18}}$

   $= 4.14 \times 10^{-11}\ \text{m} = 0.0414\ \text{nm}$.

This is a classic problem that connects two beautiful ideas: the kinetic energy gained by an electron accelerated through a potential difference, and the quantum nature of light. When an electron slams into a metal target in an X-ray tube, it can lose energy in one dramatic step — emitting a single photon. The most energetic photon possible corresponds to the electron giving up all its kinetic energy at once. That sets the upper limit on frequency and the lower limit on wavelength.

The key relationship is the de Broglie–Einstein relation for photons: $E = hf$, where $h$ is Planck’s constant. For the electron, the kinetic energy gained is $K = eV$, where $e$ is the electron charge and $V$ is the accelerating voltage. Setting $K = hf_{\text{max}}$ gives us the maximum frequency. Then $\lambda_{\text{min}} = c / f_{\text{max}}$ gives the minimum wavelength.

Let’s work through it step by step.

1. Find the kinetic energy of the electron. An electron accelerated through a potential difference $V = 30\ \text{kV} = 30 \times 10^3\ \text{V}$ gains kinetic energy

This is the maximum energy available to produce a single X-ray photon.

2. Set this equal to the photon energy for maximum frequency. The photon energy is $E = hf$. For the most energetic photon,

So

Using $h = 6.626 \times 10^{-34}\ \text{J·s}$,

(Rounding to three significant figures gives $7.24 \times 10^{18}\ \text{Hz}$ if we use $h = 6.63 \times 10^{-34}$ — both are acceptable in exams.)

3. Now find the minimum wavelength. For any electromagnetic wave, $c = f\lambda$. The minimum wavelength corresponds to the maximum frequency:

Using $c = 3.00 \times 10^8\ \text{m/s}$,

That’s $0.0414\ \text{nm}$ (since $1\ \text{nm} = 10^{-9}\ \text{m}$).

4. Check the numbers with the shortcut. From $eV = hc/\lambda_{\text{min}}$, we get

With $hc = 1240\ \text{eV·nm}$ (a very useful constant),

This confirms our calculation.

Method: De Broglie–Duane–Hunt Relation (Inverse Photoelectric Effect)

This problem uses the fact that when an electron is stopped completely in a target, its entire kinetic energy converts into a single X-ray photon. That photon has the maximum possible frequency and the minimum possible wavelength for that accelerating voltage.

Step 1 – Write the energy conversion

The kinetic energy gained by an electron accelerated through a potential difference $V$ is:

where $e = 1.6 \times 10^{-19}\ \text{C}$ and $V = 30\ \text{kV} = 30 \times 10^3\ \text{V}$.

When this electron is brought to rest in one collision, the photon produced has energy:

Step 2 – Find maximum frequency

From the equation above:

Use $h = 6.63 \times 10^{-34}\ \text{J·s}$.

Step 3 – Find minimum wavelength

Use the wave relation $c = f \lambda$:

where $c = 3 \times 10^8\ \text{m/s}$.

A useful shortcut: $hc \approx 1240\ \text{eV·nm}$ (or $1.24 \times 10^{-6}\ \text{eV·m}$). Here:

In metres:

Final Answer

1. Maximum frequency: $\mathbf{7.24 \times 10^{18}\ \text{Hz}}$
2. Minimum wavelength: $\mathbf{4.13 \times 10^{-11}\ \text{m}}$ (or $\mathbf{0.0413\ \text{nm}}$)
   Tip

   For quick calculation, remember $hc = 1240\ \text{eV·nm}$. Then $\lambda_{\text{min}}$ in nm is simply $\frac{1240}{V\ (\text{in volts})}$.

   Watch out

   Do not confuse this with the de Broglie wavelength of the electron itself. The de Broglie wavelength of a $30\ \text{keV}$ electron is about $7 \times 10^{-12}\ \text{m}$ — noticeably smaller than the X-ray photon's minimum wavelength here. They are different physical quantities.

Common Mistakes on the De Broglie / X-Ray Wavelength Problem

This question is from the X-ray production chapter, not directly from the De Broglie wavelength topic — and that itself is the first trap. Students often mix up the two concepts. Let me walk through the mistakes one by one.

Mistake 1: Using the De Broglie wavelength formula instead of the Duane–Hunt relation

The most common error: a student sees "wavelength" and "electrons" and immediately writes

This gives the De Broglie wavelength of the electron, not the X-ray wavelength. The question asks for the X-rays produced when electrons strike a target. The minimum wavelength of X-rays comes from the entire kinetic energy of the electron converting into a single photon:

How to avoid: Read the question carefully. If it says "X-rays produced by electrons," you are in the X-ray production chapter. The relevant formula is $eV = h f_{\text{max}}$ (or $eV = h c / \lambda_{\text{min}}$). The De Broglie formula belongs to a different chapter.

Mistake 2: Forgetting to convert kV to V

The voltage is given as $30\ \text{kV}$. That is $30 \times 10^3 = 3.0 \times 10^4\ \text{V}$. Students sometimes plug in 30 directly, which gives an answer off by a factor of 1000.

How to avoid: Always write the conversion explicitly: $V = 30\ \text{kV} = 30 \times 10^3\ \text{V} = 3.0 \times 10^4\ \text{V}$. Do it on paper before substituting.

Mistake 3: Using the wrong value of Planck's constant or speed of light

Two common sub-mistakes here:

• Using $h = 6.63 \times 10^{-34}\ \text{J s}$ but forgetting that $eV$ is in joules. The energy $eV$ must be in joules: $E = (1.6 \times 10^{-19})(3.0 \times 10^4) = 4.8 \times 10^{-15}\ \text{J}$.
• Using $c = 3 \times 10^8\ \text{m/s}$ but then getting the wavelength in metres — which is fine, but then you must convert to ångströms or picometres as the problem expects.

How to avoid: Keep a consistent unit system. Use SI units throughout, then convert at the end. A useful shortcut: for X-ray problems, use the formula in eV and ångströms:

This comes from $h c = 12400\ \text{eV·Å}$. For $V = 30\ \text{kV} = 30000\ \text{V}$:

Mistake 4: Confusing maximum frequency with minimum wavelength

Students sometimes calculate the frequency correctly but then write $\lambda_{\text{min}} = c / f_{\text{max}}$ and get the right answer — but they mix up which is maximum and which is minimum. The relationship is:

Since $f$ and $\lambda$ are inversely related, the maximum frequency corresponds to the minimum wavelength. There is no "maximum wavelength" in this context — the continuous X-ray spectrum has a sharp cut-off at the short-wavelength end.

How to avoid: Write the two relations side by side:

• $eV = h f_{\text{max}}$ → solve for $f_{\text{max}}$
• $eV = \frac{h c}{\lambda_{\text{min}}}$ → solve for $\lambda_{\text{min}}$

Then check: does a larger $V$ give a larger $f_{\text{max}}$? Yes. Does it give a smaller $\lambda_{\text{min}}$? Yes. That consistency check catches errors.

Mistake 5: Not showing the final answer with correct units and significant figures

Examiners expect:

• Frequency in Hz (or s$^{-1}$)
• Wavelength in metres or ångströms (often ångströms are preferred for X-rays)

For $V = 3.0 \times 10^4\ \text{V}$:

How to avoid: After calculation, ask: "Does this wavelength make sense for X-rays?" X-ray wavelengths are of the order of $10^{-10}$ to $10^{-11}$ m (0.1–1 Å). If you get $10^{-8}$ m (UV range) or $10^{-12}$ m (gamma rays), you've made an error.

Summary of the correct approach

1. Convert kV to V.
2. Use $eV$ in joules (or use the $12400\ \text{eV·Å}$ shortcut).
3. Do not use the De Broglie formula.
4. Check that your final wavelength is in the X-ray range (~0.1–1 Å).

The correct answers:

• (a) $f_{\text{max}} \approx 7.24 \times 10^{18}\ \text{Hz}$
• (b) $\lambda_{\text{min}} \approx 4.14 \times 10^{-11}\ \text{m}$ (or $0.414\ \text{Å}$)