JEE Advanced 2024 — Paper 2 — PHY-I solution

JEE Advanced2024Paper 2PHY-I

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 $K_{α}$ -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

A
$2.53$
B
$1.27$
C
$2.24$
D
$1.58$

Solution

From Moseley's law, $\frac{1}{λ _{α}} = \frac{3 R}{4} (Z - 1)^{2}$ , so $λ_{α} \propto \frac{1}{( Z - 1 ) ^{2}}$ .

The cut-off wavelength $λ_{cut} = \frac{h c}{e V}$ depends only on the accelerating voltage and is the same for both targets.

$∴ r = \frac{λ _{α}}{λ _{cut}} \propto \frac{1}{( Z - 1 ) ^{2}} .$

For $Z = 46$ , $(Z - 1) = 45$ , and $r = 2$ . For $Z = 41$ , $(Z - 1) = 40$ .

$r_{new} = 2 \times (\frac{45}{40})^{2} = 2 \times \frac{2025}{1600} \approx 2.53.$

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