A metal target with atomic number Z=46 is bombarded with a high energy electron beam. The emission o

Q: Why doesn't the cut-off wavelength change when we change the target?

The cut-off wavelength ($\lambda_{\text{cutoff}}$) is determined entirely by the kinetic energy of the incoming electron beam, not the target material. Since the problem states 'the same electron beam' is used, its energy remains constant, meaning the shortest possible wavelength it can produce remains identical.

Q: What exactly does the $Z-1$ represent in Moseley's Law?

The $Z-1$ term represents the **effective nuclear charge**. When an electron drops into the K-shell (n=1) to emit a $K_\alpha$ photon, it is shielded from the full positive charge of the nucleus ($Z$) by the one other electron already sitting in that K-shell. Hence, it 'feels' a charge of $Z-1$.

Q: Why do we square the $(Z-1)$ term?

This comes directly from the Bohr model of the atom. The energy levels of a hydrogen-like atom are proportional to $Z^2$. Since the energy of the emitted photon is the difference between two such levels, the energy (and thus the frequency) is proportional to the square of the effective nuclear charge, $(Z-1)^2$. Wavelength is inversely proportional to energy, hence the inverse square relationship.

Q: Could I have solved this by finding the exact accelerating voltage first?

Technically, yes, but it would be a massive waste of time! You would need to introduce Rydberg's constant, Planck's constant, and the speed of light, only to watch them all cancel out when you switch to the second target. Using ratios is a powerful shortcut in physics that eliminates constants.

Q: What if the question asked about the $K_\beta$ line instead of $K_\alpha$?

For the $K_\beta$ line (a transition from n=3 to n=1), the screening effect is slightly different because the electron is further away before it drops. The screening constant $b$ in Moseley's Law ($\sqrt{\nu} = a(Z-b)$) would be different (often empirically found to be around 7.4 for heavier elements), so you couldn't just use $Z-1$.

Question

A metal target with atomic number $Z=46$ is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio $r$ of the wavelengths of the $K_\alpha$ -line and the cut-off is found to be $r = 2$. If the same electron beam bombards another metal target with $Z = 41$, the value of $r$ will be

Options:

(A) 2.53

(B) 1.27

(C) 2.24

(D) 1.58

Sigma Percentile · Accepted Answer

### The Anatomy of an X-Ray Spectrum

Imagine you are standing in a high-tech physics laboratory. An electron beam, acting like a barrage of tiny, high-energy bullets, is fired at a solid metal target. When these electrons smash into the metal, a violent deceleration occurs, producing an X-ray spectrum.

This spectrum has two distinct features. First, there is a continuous background curve caused by **Bremsstrahlung** (braking radiation). The absolute shortest wavelength in this continuous spectrum is called the **cut-off wavelength** ($\lambda_{	ext{cutoff}}$). Crucially, this cut-off depends *only* on the kinetic energy of the incoming electrons (the accelerating voltage), not on the metal itself.

Second, superimposed on this continuous curve are sharp, intense spikes. These are the **characteristic X-rays**, such as the $K_\alpha$ line. These peaks occur when an incoming electron knocks out an inner-shell electron of the target atom, and a higher-shell electron drops down to fill the vacancy, emitting a photon of very specific energy.

### Moseley's Law: The Heart of the Target

Because the characteristic X-rays depend on the atomic structure of the target, their wavelengths change if we swap the metal. This relationship is beautifully captured by **Moseley's Law**.

For the $K_\alpha$ line, Moseley discovered that the frequency of the emitted X-ray is proportional to $(Z-1)^2$, where $Z$ is the atomic number. The $-1$ accounts for the screening effect of the one remaining electron in the K-shell. Since wavelength is inversely proportional to frequency, we can write:

$$\lambda_{K_\alpha} \propto \frac{1}{(Z-1)^2}$$

### Setting Up the Ratios

The brilliance of this problem lies in what remains constant. We are told that the *same* electron beam is used for both targets. This means the accelerating voltage is identical, and therefore, $\lambda_{	ext{cutoff}}$ is a constant for both scenarios.

Let's define the ratio $r = \frac{\lambda_{K_\alpha}}{\lambda_{	ext{cutoff}}}$. Since the denominator is constant, the ratio $r$ follows the exact same proportionality as the $K_\alpha$ wavelength:

$$r \propto \frac{1}{(Z-1)^2}$$

For our first target ($Z_1 = 46$), the ratio is given as $r_1 = 2$. 
For our second target ($Z_2 = 41$), the ratio is unknown, let's call it $r_2 = x$.

### The Final Calculation

To eliminate the proportionality constant, we simply divide the two cases. Because of the inverse relationship, the terms flip:

$$\frac{r_1}{r_2} = \frac{(Z_2 - 1)^2}{(Z_1 - 1)^2}$$

Now, we substitute our known values into this master equation:

$$\frac{2}{x} = \frac{(41 - 1)^2}{(46 - 1)^2}$$

$$\frac{2}{x} = \frac{40^2}{45^2}$$

Before squaring large numbers, always simplify the fraction! Both 40 and 45 are divisible by 5, reducing the fraction to $\frac{8}{9}$.

$$\frac{2}{x} = \left(\frac{8}{9}ight)^2 = \frac{64}{81}$$

Finally, we cross-multiply to solve for $x$:

$$x = \frac{2 	imes 81}{64} = \frac{162}{64}$$

$$x = 2.53125$$

Rounding to two decimal places, we get **2.53**, which perfectly matches option (A). By understanding what changes (the target) and what stays the same (the beam), a complex quantum mechanics problem reduces to elegant middle-school algebra!

Select Answer:

Visualized Solution

$λ_{cutoff} and λ_{K_{α}}$

$λ_{K_{α}} \propto (Z - 1)^{- 2}$

$Z_{1} = 46 ⟹ r_{1} = 2$

$Z_{2} = 41 ⟹ r_{2} = x$

$\frac{r _{1}}{r _{2}} = \frac{( Z _{2} - 1 ) ^{2}}{( Z _{1} - 1 ) ^{2}}$

$\frac{2}{x} = (\frac{40}{45})^{2}$

$x = 2.53$

$V \to 2 V ⟹ λ_{cutoff} \to \frac{λ _{cutoff}}{2}$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Anatomy of an X-Ray Spectrum

Moseley's Law

Setting Up the Ratios

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λcutoff​ and λKα​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λKα​​∝(Z−1)−2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z1​=46⟹r1​=2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z2​=41⟹r2​=x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2​r1​​=(Z1​−1)2(Z2​−1)2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2​=(4540​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2.53

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V→2V⟹λcutoff​→2λcutoff​​