Modern Physics 2 Question 23

22. When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA expressed in eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is TB=(TA1.50eV). If the de-Broglie wavelength of these photoelectrons is λB=2λA, then

(1994,2M)

(a) the work function of A is 2.25eV

(b) the work function of B is 4.20eV

(c) TA=2.00eV

(d) TB=2.75eV

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Solution:

  1. Kmax=EW

Therefore,

TA=4.25WATB=(TA1.50)=4.70WB

From Eqs. (i) and (ii),

WBWA=1.95eV

de-Broglie wavelength is given by

λ=h2Km or λ1KK=KE of electron 

λBλA=KAKB or 2=TATA1.5

This gives, TA=2eV

From Eq. (i) WA=4.25TA=2.25eV

From Eq. (iii) WB=WA+1.95eV=(2.25+1.95)eV

or

WB=4.20eVTB=4.70WB=4.704.20=0.50eV



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