If voltage applied to an X-ray tube is increased from V = 10 kV to 20 kV, the wave length interval between Kα -line and cut off wavelength of continuous X-ray becomes three times. Take Rhce=13.6V,hc=12420 (eV−A°), where R is Rydberg constant, h is Planck's constant, c is speed of light in vacuum and -e is charge on electron.Mark the CORRECT statement(s).

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Atomic no. of target metal used is 29

Cut-off wavelength when V = 20 kV is 1.2A0

Cut-off wavelength when V = 10 kV is 2 A0

Atomic no. of target metal used is 26

answer is A.

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Detailed Solution

Let x be the wave length for Kα -lineλcut off=hceVInitially , λcut off=hceV1=1242010000A0=1.242A0Finally , λcut off=hceV2=1242020000A0=0.621A0Given : x−hceV2x−hceV1=3.................(1)Also, 1λ=R(Z−1)2(1−122)⇒x=43R(Z−1)2.........(2)Solving (1) and (2) , we get⇒ Z=29

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