Question

To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of 0.529 ✕ 10−10 m, the average distance between proton and electron in a hydrogen atom. how much work is done by the electric field (in j)?

Answer 1

1) First of all, let's calculate the potential difference between the initial point (infinite) and the final point (d=0.529x10-10 m) of the electron.
This is given by:
$\Delta V =- \int\limits^{d}_{\infty} {E} \, dr$
Where E is the electric field generated by the proton, which is
$E=k_e \frac{q}{r^2}$ 
where $k_e=8.99\cdot10^9~Nm^2C^{-2}$ is the Coulomb constant and $q=1.6\cdot10^{-19}~C$ is the proton charge.
Replacing the electric field formula inside the integral, we obtain
$\Delta V =- \int\limits^{d}_{\infty} {k_e \frac{q}{r^2} } \, dr = k_e \frac{q}{d}= 27~V$

2) Then, we can calculate the work done by the electric field to move the electron (charge $q_e=-1.6\cdot10^{-19}C$) through this $\Delta V$. The work is given by
$W=-q_e \Delta V = - (-1.6\cdot10^{-19}C) (27V)=4.35\cdot10^{-18}~J$

Answer 2

Distance between proton and electron d = 0.529 x 10^-10 
According to Coulombs law E = k x Q / d^2
 Coulombs constant k = 9 x 10^9 Nm^2 / C^2
 Charge Q = 1.6 x 10^-19
 E =( 9 x 10^9 x 1.6 x 10^-19) / (0.529 x 10^-10)^2 = 14.4 x 10^-10 / 0.279 x 10^-20
 E = 51.61 x 10^10 J /m