Question

A energy of 6000A photon is 3.30 *10-19 find energy of of 4000A

Answer 1

We have E=hf

( f=nu)

And f=c/lambda

So E=hc/lambda

= 6.625×10*-34 × 3×10*8/4000×10^-10 ×1.6×10^-19

=3.10J