0.19 Even semester2014-midsemester examination answers  (Page 3/3)

1. This velocity can be thermal velocity or Pauli Velocity.

Answers: Conducting electron is strongly scattering in metallic crystalline lattice but very wekly scattering in Semiconductor as exhibited by short Mean Free Path in metals and by long Mean Free Paths in Semi-conductors.[1/4+1/4=1/2]

XXV. In the Table given in Question XIV which is Pauli Velocity and which is Thermal Velocity?

Answers: Conducting Electron velocities in Metals are Pauli velocities because they arise out of Pauli-Exclusion Principle whereas conducting electron velocities in Semi-conductor are thermal velocities.[1/4+1/4=1/2]

XXVI. In Question XVIII, intrinsic Ge has the least resistivity and intrinsic GaAs has the maximum , Why? Answers; Ge has the least Band-Gap and GaAs has the widest Band-Gap.[1/4+1/4=1/2]

XXVII.

XXVIII.

XXIX. If Ultra-Violet light of λ = 0.2μm is incident on a metal then which of the metals listed in the Table below will respond and emit electrons ? Answers;UV photon has 6.2eV hence it is powerful enough to cause photo-emission in all the Metals listed in the Table below.All the metals have a work-function less than 6.2eV.[1/2 Poiint]

XXX.

Question 2.Describe the step for preparing Electronic Grade Poly-crystal Silicon from sand.

[7.5 marks]

Answer:

Reduction of sand with carbon gives impure polycrystalline Silicon

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Reaction of pulverized raw silicon with HCl gaseous vapour to form TriChloroSilane

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Multiple distillation of TriChloroSilane to obtain purified electronic grade TriChloroSilane

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Thermal decomposition of SiHCl3 at 1000 degree centigrade in Sieman’s reactor to obtain fattened rods of electronic grade Silicon

Question 3. Balmer Series Spectral Lines from Stars are determined to be at 6563A°,4862A°,4341A°,4102A° and 3970A°. Determine these spectral lines theoretically.

[By the Red-Shift of these Balmer Spectral Lines the velocity of the receding Galaxy is detyermined. The recession velocity gives the distance of the Galaxy using Hubble’s relationship][7.5 marks]

Answer: λ(μm)=1.24/E g (eV) here E g = (13.6eV/4-13.6/n^ 2 )eV. Since the answer is to be obtained in Angstrom therefore the final expression is:

Therefore λ(Angstrom)=( (1.24/(13.6/4-13.6/n^ 2 ))×10^ 4 )Angstrom.

For n=3 to n=2, λ=6564.71A°;

For n=5 to n=2, λ=4862.75A°;

For n=6 to n=2, λ=4341.74A°;

For n=7 to n=2, λ=4102.94A°;

For n=8 to n=2, λ=3971.24A°;

Question 4.Detrmine the temperature at which E=E _F +kT will be occupied by electron with P(E)=1%.[7.5 marks]

Answer

This question is wrongly stated. At E=E _F +kT the Probability of occupancy is 1/(1+e) =0.2689. It can be never 1% irrespective of temperature.

The correct question is: Determine P(E _F + kT) in Fermi-Dirac Statistics. Determine Temperature T at which P(E _F + 0.5eV) = 1%.[Answers: 0.27, 1262Kelvin]

You can give either of the answers and fetch full marks.