JEE Main 17 March 2021 Shift 2 Solved Paper

We know that, the photoelectric effect equation,
KE=hf−φ
Here, KE= kinetic energy of the electrons
h= Planck's constant
f= frequency of the light and
φ= work-function of the photocathodes.
For two identical photocathodes,
‌

1

2

mv12=hf1−φ....(i)
‌

1

2

mv22=hf2−φ.....(ii)
Since, the material of photocathodes is same, so the value of the work-function φ is same.
Subtracting Eq. (ii) from Eq. (i) we get
‌

1

2

mv12−‌

1

2

mv22=hf1−hf2
v12−v22=‌

2h

m

(f1−f2)