Question

In a hydrogen atom, the electron and proton are bound at a distanceof about 0.53 Å: (a) Estimate the potential energy of the system in eV, taking thezero of the potential energy at infinite separation of the electronfrom proton. (b) What is the minimum work required to free the electron, giventhat its kinetic energy in the orbit is half the magnitude ofpotential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potentialenergy is taken at 1.06 Å separation?

Answer 1

Given: The distance between electron and proton of a hydrogen atom is 0.53  A °

a)

The potential energy at infinity is zero.

Potential energy of the system is given as,

E=0− q 1 q 2 4π ε 0 d

Where, the charge on electron is q 1 , the charge on proton is q 2 , the distance between electron-proton of a hydrogen atom is d, the permittivity of free space is ε 0 .

By substituting the given values in the above expression, we get

E=0− 9× 10 9 × ( 1.6× 10 −19 ) 2 0.53× 10 −10 =−43.7× 10 −19  J

Since, 1 eV=1.6× 10 −19  J.

By substituting the given value in the above expression, we get

E= −43.7× 10 −19 1.6× 10 −19 =−27.2 eV

Thus, the potential energy of the system is −27.2 eV

b)

The kinetic energy of the system is the half of the magnitude of the potential energy.

The kinetic energy of the system is given as,

E 1 = 1 2 ×E

By substituting the given value in the above expression, we get

E 1 = 1 2 ×( 27.2 ) =13.6 eV

The Total energy is given as,

E t = E 1 +E

By substituting the given values in the above expression, we get

E t =13.6−27.2 =−13.6 eV

Thus, the minimum work required to free the electron is 13.6 eV.

c)

Given: The distance between electron and proton of a hydrogen atom is 1.06  A ° .

Potential energy of the system is given as,

E= q 1 q 2 4π ε 0 d 1 − q 1 q 2 4π ε 0 d

Where, the distance between electron-proton of a hydrogen atom is d 1 .

By substituting the given values in the above expression, we get

E= 9× 10 9 × ( 1.6× 10 −19 ) 2 1.06× 10 −10 −27.2 eV =21.73.7× 10 −19  J−27.2 eV =13.58 eV−27.2 eV =−13.6 eV

The kinetic energy of the system is given as,

E 1 = 1 2 ×E

By substituting the given value in the above expression, we get

E 1 = 1 2 ×( 13.6 ) =6.8 eV

The Total energy is given as,

E t = E 1 +E

By substituting the given values in the above expression, we get

E t =6.8−13.6 =−6.8 eV

Thus, the potential energy is −13.6 eV and the minimum work required to free the electron is 6.8 eV.