Question

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the electric potential of the proton at the position of the electron? Express your answer with the appropriate units. nothing nothing Request Answer Part B What is the electron's potential energy? Express your answer with the appropriate units. nothing nothing Request Answer

Answer 1

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q=$1.6\times 10^{-19} C$

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U=$\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$Charge on electron

$q_p=q=1.6\times 10^{-19} C$=Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$