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In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

**(a)** Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

**(b)** What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

**(c)** What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

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#### Solution

The distance between electron-proton of a hydrogen atom, d = 0.53 Å

Charge on an electron, q_{1}_{ }= −1.6 × 10^{−19}^{ }C

Charge on a proton, q_{2}_{ }= +1.6 × 10^{−19}^{ }C

**(a)** Potential at infinity is zero.

Potential energy of the system, = Potential energy at infinity − Potential energy at distance d

= `0 - ("q"_1"q"_2)/(4piin_0"d")`

where,

∈_{0} is the permittivity of free space

`1/(4piin_0) = 9 xx 10^9 "Nm"^2 "C"^-2`

∴ Potetial energy = `0 - (9 xx 10^9 xx (1.6 xx 10^-19)^2)/(0.53 xx 10^10)`

= `-43.47 xx 10^-19 "J"`

Since `1.6 xx 10^-19 "J" = 1 "eV"`

∴ Potetial energy = `-43.7 xx 10^-19 =(-43.7 xx 10^-19)/(1.6 xx 10^-19) = -27.2 "eV"`

Therefore, the potential energy of the system is −27.2 eV.

**(b)** Kinetic energy is half of the magnitude of potential energy.

Kinetic energy = `1/2 xx (-27.2)` = 13.6 eV

Total energy = 13.6 − 27.2 = 13.6 eV

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

**(c)** When zero of potential energy is taken, `"d"_1` = 1.06 Å

∴ Potential energy of the system = Potential energy at d_{1} − Potential energy at d

= `("q"_1"q"_2)/(4piin_0"d"_1)-27.2 "eV"`

= `(9 xx 10^9 xx (1.6 xx 10^-19)^2)/(1.06 xx 10^-10)-27.2 "eV"`

= `21.73 xx 10^-19 "J" - 27.2 "eV"`

= 13.58 eV − 27.2 eV

= −13.6 eV

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