A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-BrOglie wavelength of the particle varies cyclically between two values λ1, λwith  λ1> λ Which of the following statements are true?
(a) The particle could be moving in a circular orbit with origin as centre
(b) The particle could be moving in an elliptic orbit with origin as its focus

(c) When the de-Broglie wavetength is λ1the particle is nearer the origin than when its value is λ2
(d) When the de-Broglie wavelength is λ2 the particle is nearer the origin than when its value is λ1

Answer-(b,d)

0 6 Views | Posted 3 months ago
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  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Multiple Choice Questions as classified in NCERT Exemplar

    Explanation- the debroglie wavelength of the particle can be varying cyclically between two values λ 1 and λ 2 , if particle is moving in an elliptical orbit with origin as its focus.

    Let v1, v2, be the speed of particle at A and B respectively and origin is at focus O. If λ 1 and λ 2 are the de-Broglie wavelengths associated with particle while moving at A and B

    respectively. Then,

    λ 1 =h/mv1

    λ 2  =h/mv2

    λ 1 λ 2 = v 2 v 1

    λ 1 > λ 2

    So v2>v1

    By law of conservation of angular momentum, the particle moves faster when it is closer to

    focus.

    From figure, we note that origin O is closed to P than A

     

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A
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This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

f(x)=2x23x2x2=2x24x+x2x2=2x(x2)+1(x2)x2=(2x+1)(x2)x2=2x+1limx2f(x)=2x+1=limh02(2h)+1=4+1=5limx2+f(x)=2x+1=limh02(2+h)+1=4+1=5limx2f(x)=5Aslimx2f(x)=limx2+f(x)=limx2f(x)=5Hence,f(x)iscontinuousatx=2.

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