Given curve shows the intensity wavelength relation of $X -$ rays coming from two different Coolidge tubes $A$ and $B$ . The dark curve represents the relation for the tube $A$ in which potential difference between the target and the filament is $V_{A}$ and the atomic number of the target material is $Z_{A}$ . Similarly dotted curve is for tube $B$ . Respective quantities are $V_{B}$ and $Z_{B}$ foe the tube $B$ . The,

V_{A} > V_{B}, Z_{A} > Z_{B}

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V_{A} > V_{B}, Z_{A} < Z_{B}

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V_{A} < V_{B}, Z_{A} > Z_{B}

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V_{A} < V_{B}, Z_{A} < Z_{B}

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Solution

The correct option is B $V_{A} > V_{B}, Z_{A} < Z_{B}$
$λ_{min} = \frac{h c}{e V}$
$λ \propto \frac{1}{V}$
$λ_{2} > λ_{1}$ (refer to graph)
$V_{1} > V_{2}$
$\sqrt{ν} = a (Z - b)$ (Mosley's Law)
$\begin{matrix} ν \propto (Z - 1)^{2} λ \propto \frac{1}{{(Z - 1)}^{2}} (∵ ν \propto \frac{1}{λ}) \end{matrix}$
$λ_{1} > λ_{2}$ (refer to graph for characteristic line).
$Z_{A} < Z_{B}$
$V_{A} > V_{B}$
Hence, B is correct

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Similar questions

Q. The figure shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for the tube A in which the potential difference between the target and the filament is V_A and the atomic number of the target material is Z_A. These quantities are V_B and Z_B for the other tube. Then,
(a) V_A > V_B, Z_A > Z_B
(b) V_A > V_B, Z_A < Z_B
(c) V_A < V_B, Z_A > Z_B
(d) V_A > V_B, Z_A < Z_B

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The correct option is B VA>VB,ZA<ZBλmin=hceVλ∝1Vλ2>λ1(refer to graph)V1>V2√ν=a(Z−b)(Mosley's Law)ν∝(Z−1)2λ∝1(Z−1)2(∵ν∝1λ)λ1>λ2(refer to graph for characteristic line).ZA<ZBVA>VBHence, B is correct