Question

Calculate the magnitude of the energy of the photon associated with light of wavelength 6057.8 Amstrong.

Answer 1

E=hc/lambda=6.6*10^-34*3*10^8/6057.8*10^-10

=0.0032*10^-16ev

Hope you are satisfied with the answer

PLEASE MARK AS BRAINLIEST

Answer 2

Answer:

0.0032*10^-16ev

Explanation:

PLEASE FOLLOW ME

BRAINLIESTS