11.12 Calculate the

(a) Momentum, and

(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    11.12 Potential difference, V = 56V

    Planck’s constant, h = 6.626 ×10-34 Js

    Charge of an electron, e = 1.6 ×10-19 C

    Mass of electron, m = 9.1 ×10-31 kg

    At equilibrium, the kinetic energy of each electron is equal to the acceleration potential. If v is the velocity of each electron, we can write

    12mv2 = eV

    v=2eVm = 2×1.6×10-19×569.1×10-31 = 4.44 ×106 m/s

    The momentum of each electron = mv = 9.1 ×10-31× 4.44 ×106 kgm/s = 4.04 ×10-24 kgm/s

    De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

    λ=12.27V Å = 12.2756 Å = 1.64 Å = 1.64 ×10-10 m = 0.164 nm

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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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