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+1 vote
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in Physics by (36.3k points)

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is :

(1) 3eV

(2) 2eV

(3) 4eV

(4) 1.5eV

2 Answers

+1 vote
by (33.0k points)
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Best answer

Answer is (3) 4eV

λB = 2λA

TA = 4TB ....(i)

and TB = (TA – 1.5) eV ....(ii)

from (i) and (ii)

3TB 1.5 eV 

TB = 0.5 eV 

TB = 0.5 eV = 4.5 eV – ϕB 

ϕ = 4eV

+1 vote
by (51.9k points)

Answer is (3) 4eV

Relation between De-Broglie wavelength and K.E. is 

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