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Question
Derive the expression for the energy of an electron in the atom.
Solution
Consider an electron orbiting the nucleus of an atom with the atomic number Z in the nth orbit. Let m and -e represent the electron's mass and charge, r the orbit's radius, and v the electron's linear speed.
According to Bohr's first postulate,
centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus
∴ `"mv"^2/"r" = 1/(4piε_0) = "Ze"^2/"r"^2` ...(1)
where ε0 is the permittivity of free space.
∴ Kinetic energy (KE) of the electron
`= 1/2 "mv"^2 = "Ze"^2/(8piε_0"r")` .....(2)
The electric potential due to the nucleus of charge + Ze at a point at a distance r from it is
V = `1/(4piε_0)*"Ze"/"r"`
∴ Potential energy (PE) of the electron
= charge on the electron x electric potential
= `-"e" xx 1/(4piε_0) "Ze"/"r" = - "Ze"^2/(4piε_0"r")` ....(3)
Hence, the total energy of the electron in the nth orbit is
E = KE + PE = `"-Ze"^2/(4piε_0"r") + "Ze"^2/(8piε_0"r")`
∴ E = `"-Ze"^2/(8piε_0"r")` ......(4)
This shows that the total energy of the electron in the nth orbit of the atom is inversely proportional to the radius of the orbit as Z, ε0 and e are constants. The radius of the nth orbit of the electron is
r = `(ε_0"h"^2"n"^2)/(pi"mZe"^2)` ....(5)
where his Planck's constant.
From Eqs. (4) and (5), we get,
`"E"_"n" = - "Ze"^2/(8piε_0)((pi"mZe"^2)/(ε_0"h"^"n"^2)) = - ("mZ"^2"e"^4)/(8ε_0^2"h"^2"n"^2)` ....(6)
This expresses the energy of the electron in the nth Bohr orbit. The minus sign in the expression indicates that the electron is attracted to the nucleus by electrostatic forces.
As m, Z, e, ε0 and h are constant, we get
`"E"_"n" prop 1/"n"^2`
i.e., In a stationary energy state, the energy of the electron is discontinuous and inversely proportional to the square of the primary quantum number.
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