KEmax=hu−hu0hu−hu=e×ΔvV0=ehu−ehv0 ' v ' is doubled KEmax=2hu−hu0 V0′=(ΔV)′=e2hu−ehu0 KEmaxKEmax may not be equal to 2 ⇒V0V0′ may not equal to 2 KEmax=hu−hv0 V=ehv−eh h0
This question has Statement −1 and Statement −2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1 : A metallic surface is irradiated by a monochromatic light of frequency v>v0 (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax and V0 respectively. If the frequency incident on the surface doubled, both the Kmax and V0 are also doubled. Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
Held on 30 Apr 2011 · Verified 6 Jul 2026.
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
Statement-1 is false, Statement-2 is true.
Statement-1 is true, Statement-2 is false.
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