Advertisements
Advertisements
Question
Using Bohr’s postulates, obtain the expression for total energy of the electron in the nth orbit of hydrogen atom.
Advertisements
Solution
According to Bohr’s postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit of a given radius, the centripetal force is provided by Coulomb’s force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small.
`So, (mv^2)/r = (ke^2)/r^2 => mv^2 = (ke^2)/r .................. (1)`
Where, m = mass of electron, r = radius of electronic orbit and v = velocity of electron.
`mvr = (nh)/(2pi)=> v = (nh)/(2pimr)`
From eq(1), we get that:
`m((nh)/(2pimr))^2 = (ke^2)/r => r = (n^2h^2)/(2pimr)................... (2)`
(i) Kinetic energy of electron:
`E_K = 1/2 mv^2 = (ke^2)/(2r)`
Using eq (2),we get:Ek`(ke^2)/2 (4pi^2kme^4)/(n^2h^2) = (2pi^2k^2me^4)/(n^2h^2)`
(ii) Potential energy:
`E_p = - (k(e)xx (e))/r = - (ke^2)/r`
Using eq (2), we get `E_p= -ke^2 xx (4pi^2kme^2)/(n^2h^2) = - (4pi^2k^2me^4)/(n^2h^2)`
Hence, total energy of the electron in the nth orbit
`E = E_p +E_k = (4pi^2k^2me^4)/(n^2h^2) + (2pi^2k^2me^4)/(n^2h^2) = - (2pi^2k^2me^4)/(n^2h^2) = -13.6/n^2 eV`
APPEARS IN
RELATED QUESTIONS
What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?
Explain, giving reasons, which of the following sets of quantum numbers are not possible.
- n = 0, l = 0, ml = 0, ms = + ½
- n = 1, l = 0, ml = 0, ms = – ½
- n = 1, l = 1, ml = 0, ms = + ½
- n = 2, l = 1, ml = 0, ms = – ½
- n = 3, l = 3, ml = –3, ms = + ½
- n = 3, l = 1, ml = 0, ms = + ½
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nanosecond range. If the radiation source has a duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.
Using Bohr’s postulates, obtain the expressions for (i) kinetic energy and (ii) potential energy of the electron in stationary state of hydrogen atom.
Draw the energy level diagram showing how the transitions between energy levels result in the appearance of Lymann Series.
A filter transmits only the radiation of wavelength greater than 440 nm. Radiation from a hydrogen-discharge tube goes through such a filter and is incident on a metal of work function 2.0 eV. Find the stopping potential which can stop the photoelectrons.
Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of the hydrogen atom. The ground state energy of the hydrogen atom is −13.6 eV.
Ratio of longest to shortest wavelength in Balmer series is ______.
When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula
`bar(v) = 109677 1/n_1^2 - 1/n_f^2`
What points of Bohr’s model of an atom can be used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term.
The wavelength of the first time line of Ballmer series is 6563 A°. The Rydberg constant for hydrogen is about:-
The ground state energy of hydrogen atoms is -13.6 eV. The photon emitted during the transition of electron from n = 3 to n = 1 unknown work function. The photoelectrons are emitted from the material with a maximum kinetic energy of 9 eV. Calculate the threshold wavelength of the material used.
