In Bohr's model of Hydrogen atom, what is the total energy of the electron in n-th discrete orbit
Answers
Answer:
Consider the electron revolving in the nth orbit around the hydrogen nucleus. Let m and -e be the mass and the charge of the electron, r the radius of th eorbit and v, the liner speed of the v, the linear speed of the electron.
According to Bohr's first postulate, centripetal force acting on the electron = electrostatic force of attraction exerted on the electron by the nucleus
∴mv2r=14πε0⋅e2r2 ... (1)
where ε0 is the permittivity of free space.
∴ Kinetic energy (KE) of the electron =12mv2=e28πεr ...(2)
The electric potential due to the hydrogen nucleus (charge = +e) at a point at a distance r from it is V=14πε0⋅er
∴ Potential energy (PE) of the electron
=charge on the electron × electric potential
=−e×14πε0er=−e24πε0r ...(3)
Hence, the total energy of the electron in the nth orbit is
E=KE+PE=−e24πε0r+e28πε0r
∴E=−e28πε0r ...(4)
This shows that the total energy of the electron in the nth orbit of hydrogen atom is inversely proportional to the radius of the orbit as ε0 and e are constants.
The radius of the nth orbit of the electron is
r=ε0h2n2πme2 ...(5)
where h is Planck's constant.
From Eqs. (4) and (5), the energy of the electron in the nth Bohr orbit is
En=−e28πε0(πme2ε0h2n2)
=me48ε20h2n2 ...(6)
The minus sign in the shows that the electron is bound to the nucleus by the electrostatic force of attraction.
As m, e, ε0 and h are constant, we get,
En∝1n2
i.e., the energy of the electron in a stationary energy state is discrete and is inversely proportional to the square of the principal quantum number.
Explanation:
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