X-rays and Moseley's Law JEE 2024: Wavelength Ratio

What changes when the same electron beam hits a different target?

The cut-off wavelength stays fixed; the characteristic wavelength increases, giving option A, 2.53. This X-rays and Moseley's law JEE 2024 question is from JEE Advanced 2024, Paper 2, Physics, Atoms and Nuclei. It is a single-correct MCQ, classified as medium by the question bank, not by the examination authority.

A high-energy electron beam bombards a metal target of atomic number 46. Its K-alpha to cut-off wavelength ratio is:

The unchanged beam then bombards a target of atomic number 41. Find the new ratio from the supplied options:

Only the wavelength dependences are needed here. No apparatus geometry needs reconstruction.

Which wavelength depends on atomic number, and which depends on the beam?

Moseley’s law gives the characteristic K-alpha frequency as:

The characteristic wavelength depends on the target’s atomic number. The cut-off wavelength depends on the incident electron’s energy, so it stays fixed for the same beam. The official solution separates these two dependences before forming the ratio.

For electromagnetic radiation: $\lambda =\frac{c}{\nu }$

Since the speed of light is constant, dividing by a larger frequency gives a smaller wavelength. Thus, the frequency dependence becomes an inverse wavelength dependence:

The subtraction of one represents the screening approximation used for this K-alpha relation. Another electron partly screens the nuclear charge, so use the screened factor, not the atomic number alone.

The cut-off wavelength is:

Here, the symbols denote Planck’s constant, the speed of light, elementary charge and accelerating voltage, respectively. The product of elementary charge and accelerating voltage is the incident electron’s kinetic energy.

The cut-off corresponds to the maximum photon energy available from one incident electron. It depends on beam energy, not target atomic number.

The unchanged beam means unchanged accelerating voltage, so the cut-off remains fixed. Combining the dependences gives:

How does the official calculation give option A, 2.53?

Multiplying the original ratio by the increase in characteristic wavelength gives 2.53, option A. In the official proportionality method, put the old target’s ratio in the numerator on the left and its wavelength dependence in the numerator on the right. Matching that order avoids reversing the result.

Retain the original ratio and label the unknown:

Using the inverse-square dependence:

Subtract one from each atomic number and simplify the division:

Rearrange:

Square and evaluate:

The correct answer is option A. Decreasing atomic number lowers the K-alpha frequency and increases its wavelength. Since the cut-off wavelength is fixed, the new ratio must exceed the original value of 2.

Neither the accelerating voltage nor either absolute wavelength is needed. The unchanged beam factor and the common characteristic-line proportionality constant cancel when the two ratios are compared.

Why does option B, 1.27, appear in the calculation?

Option B is the multiplier, not the new ratio. Stopping after calculating the change factor, before applying it to the initial value, gives:

This is the new ratio divided by the old ratio:

It measures the factor by which the ratio increases. It does not give the new ratio itself, because the starting value was 2, not 1.

Restore that missing initial value:

Label the intermediate result “multiplier”, then multiply by the initial ratio. Keep that label in your working rather than leaving an isolated decimal.

The earlier direction check also rejects 1.27: it is below 2. The characteristic wavelength has increased while the denominator has stayed fixed, so the final ratio cannot decrease.

How do you apply these dependences to related practice questions?

Track what changes: the target sets the characteristic frequency, while the beam energy sets the cut-off wavelength. Check the order of comparison before forming a ratio. These three questions are original practice variants derived from the solved question, not additional verified PYQs.

What is the K-alpha frequency ratio for atomic numbers 41 and 46 with the same beam?

Practice variant 1: Use Moseley’s direct-square dependence because the question asks for frequency, not wavelength. The answer is:

The result is below one, as required for the lower-atomic-number target. Using the wavelength multiplier here would reverse the dependence.

What happens to the original ratio if the voltage doubles but the target stays unchanged?

Practice variant 2: the new ratio is 4. Start with the original target of atomic number 46 and ratio 2. Retaining the target keeps its characteristic wavelength fixed, assuming K-alpha emission remains possible.

Doubling the voltage halves the cut-off wavelength:

The same numerator divided by half the denominator gives:

What happens when the target changes back from atomic number 41 to 46?

Practice variant 3: the ratio returns to 2. Start with the unrounded ratio 2.53125 at atomic number 41, then use the same beam at atomic number 46. The inverse-square dependence gives a shorter characteristic wavelength:

Keep the unrounded starting value to recover the original ratio exactly. Next, practise a different topic within the same chapter: Atoms and Nuclei JEE 2024: Uranium–Lead Age Calculation.