Question

Consider a silver target in coolidge tube to produce x-rays. The accelerating potential is 31 kV. `E_(K)=25.51 KeV, E_(L)=3.5 1 KeV. "If" lambda_(K alpha)-lambda_("min")` is approximately8N pm(in pm), where N is an integer find N. Round off to nearest integer.
(Take : hc = 1240 e V nm)

Answer 1

Correct Answer - `0002`
`lambda_("min")=(1240)/(31xx10^(3))=40xx10^(-2) nm = 4xx10^(-11)m=40 `pm
`E_(k)-E_(L)=(hc)/(lambda_(k alpha))=25.51-3.5`
`lambda_(k alpha)=(1240)/(22)xx10^(-3)=56.36` pm `=16.36 ~=16`pm