# If accelerating potential increases from20kv to 80kv in an electron microscope, its resolving power R would change to

Arun
25758 Points
4 years ago
Dear Sayam

$\lambda$ = h / P = (h / $\small \sqrt$2me)(1 / $\small \sqrt$v)
R.P. $\small \alpha$ 1 / $\lambda$  $\small \alpha$  $\small \sqrt$v
R / R’ = $\small \sqrt$(20/ 80) = 1/ 2
R’ = 2R
hope it helps
Regards