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Grade: 12
        
When a photon of energy 4.25eV strike the surface of metal A,the ejected photoelectrons  have maximum kinetic energy TA (means A to subscript of T) and debroglie wavelength LA(lamda A). 
The maximum kinetic energy of photoelectrons  liberated from another metal B by photon of energy 4.70eV is TB where TB=(TA-1.50)eV. If the debroglie wavelength 
of these photoelectrons is LB=2LA (lamda(B)=2.lamda(A)),then 
A)The work function of A is 2.25eV.    B)The work function of B is 4.20eV
C)TA=2.00eV.            D)TB=2.75 eV
It is MORE THAN ONE CORRECT OPTION TYPE QUESTION. I want a detailed solution of this PLEASE
2 years ago

Answers : (1)

jitender
114 Points
							Using formula, ¥=h/√(2mE)¥=wavelength h= plancs constantm=mass of eE=kinetic energy.Now ...TAKING RATIO ...and after cancelling same terms¥A/ ¥B=√Tb/taPutting ¥b=2¥a (given)1/2=√Tb/TaSquaring both sides.1/4= Tb/TaTa=4TbPut Tb=Ta-1.5 (given)Now Ta=2eV and Tb= 0.5Then From equationTa=Ep-☆a☆=work functionEp=energy of photon.☆a=2.25Simillarly☆b=4.2 from Tb=Ep-☆bSo a,b,C options are correct
						
2 years ago
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