If the short wavelength limit of the continous spectrum coming out of a Coolidge tube is `10 A`, then the de Broglie wavelength of the electrons reaching the target netal in the Coolidge tube is approximately

Options

A. 0.3 Å
B. 3Å
C. 30Å
D. 10Å

Answer

A

Explanation

We have K.E.`=(P^(2))/(2m_(0)) =(hc)/(lambda_("min"))`
`rArr P=sqrt((2hcm_(0))/(lambda_("min")))`
Also `lambda_("de broglie")=h/p=sqrt((hlambda_("min"))/(2m_(0)C))`
For `lambda_("min")=10Å: lambda_("de broglie")=0.3 Å 0`