Q. 264.0( 53 Votes )
Derive an express
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.
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:
From equation (1) and (3) we get,
For nth orbit,
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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