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Question
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?
Solution
The distance between electron-proton of a hydrogen atom, d = 0.53 Å
Charge on an electron, q1 = −1.6 × 10−19 C
Charge on a proton, q2 = +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 d1 − 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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