TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Given that, work-function, φ0=1.5eV
Let λ0= threshold wavelength.
Then, ‌‌φ0=‌

hc

λ0

⇒1.5eV‌‌=‌

1242

λ0

(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.