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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
  1. v0 for metal A is greater than that for metal B
  2. v0 for metal A is less than that for metal B
  3. Kmax for metal A is less than that for metal B
  4. 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
Solution
Verified by Toppr

ϕA>ϕB
ϕ=hV0
V0)A>V0)B
Kmax=hVhV0
Kmax)A<Kmax)B

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