May 2011 - Career Point

(A)MomentumMomentumPositionhas an angular frequency ω, where a part of theenergy is absorbed and a part of it is reflected. Asω approaches ω p , all the free electrons are set toresonance together and all the energy is reflected.This is the explanation of high reflectivity ofmetals.38. Taking the electronic charge as 'e' and thepermittivity as 'ε 0 ', use dimensional analysis todetermine the correct expression for ω p .(A)Nemε 0(B)mε 0Ne(B)MomentumPositionAns.Sol.(C)[C]Nemε20F = mω 2 l ≡e24πεl02(D)mεNe02ω 2 ≡e24πεl03≡⎛ e⎜⎝ m2ε 0⎞⎟⎠⎛ Nl⎜3⎝ l3⎞⎟⎠(C)Positionω =Nemε20Ans.[B](D)MomentumPositionParagraph for Questions 38 to 39A dense collection of equal number of electronsand positive ions is called neutral plasma. Certainsolids containing fixed positive ions surroundedby free electrons can be treated as neutral plasma.Let 'N' be the number density of free electrons,each of mass 'm'. When the electrons aresubjected to an electric field, they are displacedrelatively away from the heavy positive ions. Ifthe electric field becomes zero, the electronsbegin to oscillate about the positive ions with anatural angular frequency 'ω p ', which is called theplasma frequency. To sustain the oscillations, atime varying electric field needs to be applied that39. Estimate the wavelength at which plasmareflection will occur for metal having the densityof electrons N ≈ 4 × 10 27 m –3 . Take ε 0 = 10 –11and m ≈ 10 –30 , where these quantities are inproper SI unit-(A) 800 nm(B) 600 nm(C) 300 nm(D) 200 nmAns. [B]c = λf22πc Neω p = ω = =λ mελ = 2πcmεNe02=8πce0mεN2 0−302×3.14×3×10 (10 )(10=−19271.6×10 4×10= 589 × 10 –9 m ≈ 600 nm−11SECTION – IVNumerical Response TypeThis section contains 7 questions. The answer to eachquestion is a single-digit integer, ranging from 0 to 9.The bubble corresponding to the correct answer is tobe darkened in the ORS.)XtraEdge for IIT-JEE 71 MAY 2011