Question

The ratio of kinetic energy to the total energy of an electron in a bohr orbit of the hydrogen atom is

Answer 1

According to Bohr's theory,
Electrostatic force = centripetal force
K(Ze)e/r² = mv²/r
mv² = K(Ze)(e)/r
∴ Kinetic energy = 1/2 mv² = K(Ze)(e)/2r
And potential energy = F.dr = -K(Ze)(e)/r
so, Total energy = P.E + K.E = -K(Ze)(e)/2r

Kinetic energy of an electron in an orbit , K.E = K(Ze)(e)/2r
Total energy of an electron in an orbit , T.E = -K(Ze)(e)/2r

Hence, Ratio of K.E and T.E ={K(Ze)(e)/2r}/{-K(Ze)(e)/2r} = -1/1
∴ answer = -1 : 1

Answer 2

Based on Bohr's theory, electrostatic force and centripetal forces are equal. So
K(Ze)(e)/r² = mv²/r → mv² = K(Ze)(e)/r → mv²/2 = K(Ze)(e)/2r
Kinetic energy = mv²/2 = K(Ze)(e)/2r
Potential energy = -K(Ze)(e)/r
Total energy = KE + PE = -K(Ze)(e)/2r

Hence, the ratio of kinetic energy and total energy would be 1:-1