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The wavelength for $k_{β}$ for x-ray of certain material A is 12.42 pm. It takes 10 keV to remove the electron from M shell of A. The minimum accelerating potential that should be applied across x-ray tube with target material A, so that a K_a x-ray will be produced, is

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10 kV

100 kV

110 kV

80 kV

answer is C.

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Detailed Solution

${ΔE}_{K_{β}} = E_{K} - E_{M} \Rightarrow \frac{hc}{λ_{K_{β}}} = E_{K} - E_{M}$

$\Rightarrow \frac{12420 eV \overset{0}{A}}{12.42 \times 10^{- 12}} = E_{K} - 10 keV$

$∴ E_{K} - 10 keV = 100 keV \Rightarrow E_{K} = 110 keV$

${eV}_{pd} = 110 keV \Rightarrow V_{pd} = 110 kV$

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