The work function of metal 'A' is greater than that for metal B. The two metals are illuminated with appropriate radiation of frequency v so as to cause photoelectric emission in both metals. If v0 is the threshold frequency and Kmax is the maximum kinetic energy of photoelectrons, then
v0 for metal A is greater than that for metal B
v0 for metal A is less than that for metal B
Kmax for metal A is less than that for metal B
Kmax for metal A is greater than that for metal B
A
Kmax for metal A is greater than that for metal B
B
Kmax for metal A is less than that for metal B
C
v0 for metal A is greater than that for metal B
D
v0 for metal A is less than that for metal B
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Solution
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ϕA>ϕB ϕ=hV0 V0)A>V0)B Kmax=hV−hV0 ∴Kmax)A<Kmax)B
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