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Using the Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.
Let v be the velocity of the electron
Number of revolutions per unit time = f = 2πa0/v
The electric current is given as I = Q
When change Q flows in time T
Such that Q = e
Therefore, the electric current is given as I = 2πa0/v e
Show that the first few frequencies of light that are emitted when electrons fall to the nth level from levels higher than n are approximate harmonics (i.e. in the ratio 1 : 2: 3…) when n >>1.
Frequency of any line in the series on the spectrum of hydrogen is expressed as the difference between the two atoms
fcm = cRZ2[1/(n+p)2 – 1/n2]
where m = n + p
Substituting for p we get,
fcm = (2cRZ2/n3)p
What is the minimum energy that must be given to an H atom in ground state so that it can emit an Hγ line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such Hγ photon?
Energy required for the transition from n = 1 to n = 5 is given as:
E = E5 – E1 = (-13.6/52) – (-13.6/12) = 13.06 eV
Angular momentum is given as = change in angular momentum of electron = 5(h/2π) – 2(h/2 π) = 3h/2 π = 3.17 × 10-34 Js.
Long AnswersThe first four spectral lines in the Lyman series of an H-atom are λ = 1218 Å, 1028Å, 974.3 Å and 951.4Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.
Let μH and μD be the reduced masses of electrons of hydrogen and deuterium respectively.
ni and nf are the fixed mass series for hydrogen and deuterium
Rh/Rd = μH/ μD
Reduced mass of hydrogen = μH = me/1+me/M = me (1-me/M)
Reduced mass of deuterium = μD = 2M.me/2M(1+me/2M) = me(1- me/2M)
μH/ μD = 0.99973
λD/ λH = 0.99973
λD = 0.99973 λH
Therefore, change in wavelength = λH – λD = 0.3 Å
Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in 1H and 2H. This is because the wavelength of transition depends to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass µ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here µ = me M/(me +M) where M is the nuclear mass and me is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in 1H and 2H. (Mass of 1H nucleus is 1.6725 × 10–27 kg, Mass of 2H nucleus is 3.3374 × 10–27 kg, Mass of electron = 9.109 × 10–31 kg.)
The energy of an electron in the nth state is
En = -μZ2e4/8ε02h2(1/n2)
For hydrogen atom, μH = meMh/(me + Mh)
For deuterium atom, μD = meMd/(me + Md)
The wavelength of the transition for hydrogen is = λH = 3/4 μHe4/8ε02h2c
The wavelength of the transition for deuterium is = λD = 3/4 μDe4/8ε02h2c
The difference between the wavelength is = λ = λH – λD = 2.714 × 10-2%
If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1Å, and (ii) R = 10 Å.
The kinetic energy of the H atom is 13.6 eV
Potential energy of an electron and proton is -27.2 eV
Total energy of the electron = -13.6 eV
When R = 0.1Å, the ground state energy is same as the earlier = -13.6 eV
When R = 10 Å,
The kinetic energy = 0.16 eV
The potential energy of electron and proton = -3.83 eV
Total energy of the electron = -3.67 eV.
In the Auger process, an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from an = 2 to n = 1 transition.
The energy in the nth state = -Rch Z2/n2 = -13.6 Z2/n2 eV
Where R is the Rydberg constant and Z = 24
Energy released is (13.6Z2)(3/4)
Energy required to eject 4 electron is 13.6 Z2/16 eV
Therefore, the kinetic energy is 13.6Z2(3/4 – 1/16) = 5385.6 eV.
The inverse square law in electrostatics is
for the force between an electron and a proton. The 1 r dependence of | F | can be understood in quantum theory as being due to the fact that the ‘particle’ of light (photon) is massless. If photons had a mass mp, force would be modified to
where λ = m cp /ℏ and ℏ = ℏ/2π . Estimate the change in the ground state energy of an H-atom if mp were 10–6 times the mass of an electron.
Mass of photon = 9.1 × 10-37 kg
Wavelength associated with the photon = h/mpc
Total energy E = -13.6 + 27.2 λ rA
Where λrA = δ
The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge + q1, –q2 is modified to
Calculate in such a case, the ground state energy of an H-atom, if ε = 0.1, R0 = 1Å.
Case 1: When ε = 0.1, R0 = 1Å
R1 = 8 × 10-11
M = 0.08 nm
Velocity at ground level is v1 = 1.44 × 106 m/s
Case 2:
When n = 1 and v1 = h/mr1
Kinetic energy = 5.9 eV
Potential energy = -17.3 eV
Total energy = -11.4 eV
The basic unit of matter, which is the smallest thing in existence, is called an atom. The size of an atom is usually variable in nature, depending on the type of element. Protons, neutrons and electrons are the three different particles present inside the atom. Atoms, when grouped together, form molecules. For example, an oxygen molecule is made from smaller oxygen atoms which are grouped together.
The different parts of an atom are