In a Coolidge tithe, the potential difference used to accelerate the electrons is increased from 24.8 kV to 49.6 kV. As a result, the difference between the wavelength of $K_{α}$ -line and minimum wavelength becomes two times. The initial wavelength of the $K_{α}$ -line is $[Take \frac{hc}{e} - 12 .4 {kVA}^{0}]$

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$\frac{3}{2} A^{0}$

$\frac{5}{4} A^{0}$

$\frac{5}{2} A^{0}$

$\frac{3}{4} A^{0}$

answer is B.

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Detailed Solution

$Let ΔE be the energy of k_{α} -line then \frac{hc}{λ_{K_{α}}} = {ΔE}_{K_{α}} \Rightarrow λ_{K_{α}} = \frac{hc}{{ΔE}_{K_{α}}}$
Now the cutoff wavelength in the two cases are

$λ_{1_{\min}} = \frac{hc}{e \times 24 .8 kV} = \frac{12 .4}{24 .8} A^{0} = \frac{1}{2} A^{0}$
$\begin{array}{l} λ_{2_{min}} = \frac{hc}{e \times 49 .6 kV} = \frac{12 .4}{49 .6} A^{0} = \frac{1}{4} A^{0} \\ \Rightarrow 2 (λ_{K_{α}} - 1 / 2) = (λ_{K_{α}} - 1 / 4) \Rightarrow 2 λ_{K_{α}} - 1 = λ_{K_{α}} - 1 / 4 \\ \Rightarrow λ_{K_{α}} = 1 - \frac{1}{4} \Rightarrow λ_{K_{α}} = \frac{3}{4} A^{0} \end{array}$