11.2 The work function of cesium metal is 2.14 eV. When light of frequency 6 *1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the

(a) Maximum kinetic energy of the emitted electrons,

(b) Stopping potential, and

(c) Maximum speed of the emitted photoelectrons?

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8 months ago

11.2 Work function of cesium metal, 0 = 2.14 eV

Frequency of light, ν = 6 *1014 Hz

The maximum kinetic energy is given by the photoelectric effect, K = hν-0 ,

Where h = Planck's constant = 6.626 *10-34 Js, 1 eV = 1.602 *10-19 J

K = 6.626*10-34*6*10141.602*10-19-2.14 = 0.345 eV

Hencethemaximumkineticenergyoftheemittedelectronis0.3416eV

For stopping potential V0 , we can write the equation for kinetic energy as:

K = V0 , where e

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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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