1.

What is de-Broglie wavelength of electron having energy 10 KeV ?

Answer»

0.12 `Å`
1.2 `Å`
12.2 `Å`
NONE of these

Solution :`LAMBDA = (12.3)/(SQRT(v)) Å = (12.3)/(sqrt(10 xx 10^(3))) = 0.12 Å`


Discussion

No Comment Found

Related InterviewSolutions