Given that, work-function,
φ0=1.5eV Let
λ0= threshold wavelength.
Then,
‌‌φ0=‌ ⇒1.5eV‌‌=‌(eV−nm) λ0‌‌=828nm Now, photocurrent is directly proportional to intensity of incident light.
V1,V2,V3 and
V4 are cut off voltage, which is required to stop the electron having maximum kinetic energy.
Given that,
λA=200nm, its corresponding cut off voltage is
V2.
λB=400nm, its corresponding cut off voltage is
V3.
λC=600nm, its corresponding cut off voltage is
V4. and corresponding photocurrent,
IA‌‌=1.8W∕m2→III IB‌‌=1W∕m2→II IC‌‌=0.5W∕m2→IV Hence, consequently option (d) is correct.