Question

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

Answer 1

Answer:

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

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Comments Report

vishesh180999

vishesh180999 Ambitious

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

∴ answer = -1 : 1