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If the short wavelength limit of the continuous spectrum coming out of a Coolidge tube is 10 Ao, then the de Broglie wavelength of the electrons reaching the target metal in the Coolidge tube is approximately

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a

0.3Ao

b

3 Ao

c

30 Ao

d

10 Ao

answer is A.

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Detailed Solution

We have KE=P22me=hcλminP=2hcmeλmin Also, λde Broglie =hp=hλmin2mec For λmin=10Å, c=3×108m/s , h=6.6×10-34Js, me =9.1×10-31kg we get λde Broglie ≅0.3Å

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