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Q.

The wavelength for kβ 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 Ka x-ray will be produced, is

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a

110 kV

b

10 kV

c

100 kV

d

80 kV

answer is C.

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Detailed Solution

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ΔEKβ=EK−EM⇒hcλKβ=EK−EM

⇒12420 eVA012.42×10−12=EK−10 keV

∴EK−10keV=100keV⇒EK=110keV

eVpd=110 keV⇒Vpd=110 kV

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