11.13 What is the

(a) Momentum,

(b) Speed, and

(c) De Broglie wavelength of an electron with kinetic energy of 120 eV.

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1 Answer
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8 months ago

11.13 Kinetic energy of electron, Ek= 120 eV = 120 *1.6*10-19 J

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

The kinetic energy of electron can be written as Ek=12 m v2

v = 2Ekm = 2*120*1.6*10-199.1*10-31 = 6.496 *106 m/s

Momentum of the electron, p = mv = 9.1 *10-31* 6.496 *106 = 5.911 *10-24 kgm/s

Speed of the electron =

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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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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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