Question

derive an expression for the radius of the first orbit of the electron of hydrogen atom​

Answer 1

According to Bohr’s model, the electron revolves revolve in stationary orbits where the angular momentum of electron is an integral multiple of h/2π.

_______________ (1)

Here, h is Planck's constant.

Now, when an electron jumps from an orbit of higher energy E2 to an orbit of lower energy E1, it emits a photon. The energy of the photon is E2-E1. The relation between wavelength of the emitted radiation and energy of photon is given by the Einstein - Planck equation.

E2-E1= hf ____________ (2)

For an electron of hydrogen moving with a constant speed v along a circle of radius R with the center at the nucleus, the force acting on the electron according to Coulomb’s law is:

The acceleration of the electron is given by v2/r. If m is the mass of the electron, then according to Newton’s law:

F=ma

_______________ (3)

From equation (1) and (3) we get,

For nth orbit,

OR

The K.E is equal to the (-) ve of the energy of the electron and P.E is equal to twice of the energy of the electron.

K.E in ground state =

P.E in ground state =2× (-13.6) =-27.2eV

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Answer 2

The expression of radius for first orbit of H atom is rn = h^2 / 4π^2mke^2

Explanation:

When the electron revolves in the orbit it has centripetal force and coulomb force acting on it.

F centripetal = F coulomb

mv^2 / rn = ke^2 / rn^2

Where K = 1/ 4πϵ o  = 9 x 10^9

Also we know that

mvnrn = nh / 2 π

After eliminating n we get.

rn = ke^2 / m . 4π^2m^2 r^2 n / n^2h^2

rn = h^2n^2 / 4π^2mke^2

For n = 1

The equation would be

rn = h^2 / 4π^2mke^2

General expression of radius is rn = 0.529 n^2 A  o

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