Question

A proton is located at a distance of 0.050 m from a point charge of +8.30 ?C. The repulsive electric force moves the proton until it is at a distance of 0.15 m from the charge. Suppose that the electric potential energy lost by the system were carried off by a photon. What would be its wavelength? answer isn't 1.248e-4nm

Answer 1

Potential energy lost is given by:

dPE = PE2 - PE1

dPE = kq1q2/r2 - kq1q2/r1

q1 = 8.30 uC

q2 = 1.6*10^-19 C

r2 = 0.15 m

r1 = 0.05 m

So

dPE = k*q1*q2*(1/r2 - 1/r1)

dPE = 9*10^9*1.6*10^-19*8.30*10^-6*(1/0.15 - 1/0.05)

dPE = -1.59*10^-13 J (-ve sign means energy lost)

Energy lost = 1.59*10^-13 J

wavelength of photon is given by:

E = hc/lambda

lambda = hc/E

lambda = 6.626*10^-34*3*10^8/(1.59*10^-13)

lambda = 1.25*10^-12 m = 1.25e-3 nm

Please Upvote.

See that in my answer there is 'e-3' in place of 'e-4'