Question

Given light has a wavelength of 550 nm. through what potential difference must be an electron be accelerated to have this wavelength?

Answer 1

Answer:

4.9 x 10-6 Volts

Explanation:

QV = (KE)electron = (mv2)/2 = p2/2m

since p = mv

But lambda = h/p

therefore, p = h/(lambda)

so that, V = p2/(2mQ) = h2/(2mQ(lambda)2) = (6.6 x 10-34)2/(2 x 9.1 x 10-31 x 1.6 x 10-19 x (550 x 10-9)2)

V = 4.9 x 10-6 Volts