Question
In the Bohr model of the hydrogen atom, the electron is inorbit about the nuclear proton at a radius of 5.3*10^-11 m. a. use Gauss' Law to determine the electric field at apoint radially 1.00cm away from the proton b. use Gauss' Law to determine the electric field ata point radially 0.20 angstroms away from theproton. (1Angstrom = 1*10^-10 m and the radius of theproton is 10^-15 m) c. Determine the magnitude and direction of the Coulomb'sForce exterted on the electron by the proton d. Determine how fast the electron is moving around thenucleus. [centripetal force is F=(mv^2)/r ]
Explanation / Answer
The electron is in orbit about the nuclear proton at a radiusof r = 5.3 * 10-11 m a.According to Gauss's law,the net flux through any closedsurface surrounding a point charge q is given by(q/o) and is independent of the shape of thatsurface.Therefore,we get E= (q/o)--------------(1) Here,E is the electric flux and is defined asthe product of the magnitude of the electric field E and surfacearea A perpendicular to the field and is given by E= E * A ----------------(2) From equations (1) and (2),we get E * A = (q/o) Here,A is the surface area of the sphere where the electricfield is to be calculated,q is the charge on the proton ando is the permitivity of free space. or E * 4r12=(q/o) or E = (1/4r12) *(q/o) or E = (1/4o) *(q/r12) ---------------(3) or E = (1/4o) *(q/r12) ---------------(3) Here,(1/4o) = 9 * 109Nm2/C2,q = +e = +1.6 * 10-19C and r1= 1.00 cm = 1.00 * 10-2m, Substituting the above values in equation (3),we get E = 9 * 109 * (1.6 * 10-19/(1.00 *10-2)2) or E = 14.4 * 10-6 N/C b.The electric field at a point radially 0.20 angstroms awayfrom the proton is E1= (1/4o) *(q/r22) Here,r2= 0.20 angstroms = 0.20 * 10-10m or E1= 9 * 109 * (1.6 *10-19/(0.20 * 10-10)2) or E1= 36 * 1011 N/C c.The magnitude of the Coulomb's force exerted on the electronby the proton is F = (1/4o) *(q1q2/r2)----------------(4) Here,q1 and q2 are the charges on theelectron and proton respectively and r is the distance between thetwo charges. Here,q1= -1.6 * 10-19 C,q2=1.6 * 10-19 C and r = 5.3 * 10-11 m Substituting the above values in equation (4),we get F = 9 * 109 * (-1.6 * 10-19 * 1.6 *10-19/(5.3 * 10-11)2) or F = -0.82 * 10-7 N The negative value of force indicates that the force betweenthe electron and the proton is attractive.The direction of theforce is towards the center of the orbit. d.Let the speed of the electron around the nucleus bev.Therefore,we get F = (me * v2/r) ---------------(5) Here,me is the mass of the electron and r is theradius of the electron orbit. From equation (5),we get v2= (F * r/me) or v = (F * r/me)1/2 Here,me= 9.1 * 10-31 kg or v = (0.82 * 10-7 * 5.3 * 10-11/9.1 *10-31)1/2 or v = 2.18 * 106 m/s or v = 2.18 * 106 m/s