Question

A proton is located at a distance of 0.049 m from a point charge of +8.30 ?C. The repulsive electric force moves the proton until it is at a distance of 0.16 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 1

Charge on proton, q1 = + 1.602 x10^-19 C

Charge on the point charge, q2 = +8.30 C

initial distance between the charges, r1 = 0.049 m

Final distance between the charges, r2 = 0.16 m

Distance moved by proton due to repulsive electric force = 0.16 m - 0.049 m = 0.111 m

Initial electric potential energy, U1 = (Kq1xq2) / r1

Where K is coulombs constant and K = 1/4₀ = 9 x 10⁹ Nm² /C²

Hence U1 = (Kq1xq2) / r1 = (9 x 10⁹ Nm² /C² )x( + 1.602 x10^-19 C)x(+8.30 C) / (0.049m)

= 2.44 x 10^-7 J

Final electric potential energy, U2 = (Kq1xq2) / r2

Where K is coulombs constant and K = 1/4₀ = 9 x 10⁹ Nm² /C²

Hence U2 = (Kq1xq2) / r2 = (9 x 10⁹ Nm² /C² )x( + 1.602 x10^-19 C)x(+8.30 C) / (0.16m)

= 7.48 x 10^-8 J

Hence potential energy lost = 7.48 x 10^-8 J - 2.44 x 10^-7 J = - 16.92 x10^-8 J = - 1.692 x 10^-7 J

Now this lost in potential energy is carried off by the proton.

Hence energy associated with proton, E = 1.692 x 10^-7 J = hC/

=> (wavelength) = hC/E

where h = Planck's constant = 6.626x10^-34 Js

C = velocity of light = 3x10⁸ m/s

=> (wavelength) = hC/E = (6.626x10^-34 Js)x(3x10⁸ m/s )/ (1.692 x 10^-7 J ) =1.175 x10^-18 m