JEE Advanced Model Paper 8 with solutions for online practice

Energy of Lβ a c ray is

ke

λ

=

1242

15.53

kev = 80 kev
∴ Em - Ek 80 kem ..... (1)
It is given that 10 kev energy is required to remove an electron from the M – shell, so that
E0−Em = 10 kev ..... (2)
From (1) and (2) we get E∞ - Ek 90 kev
Since E∞ = 0 , Ek = - 90 kev . Thus 90 kev kev energy is required to knock out an electron from the kshell. Hence the minimum acceleration potential is 9 × 104 V . If power of 10−4 V