JEE Advanced 2016 Paper 1

For photoelectric effect, we have the relation ‌

hc

λ

=φ+eV0
where hc∕λ is the energy of the incident photon, φ is the work function and eV0 is the kinetic energy of the electron emitted.
From the given data, for different wavelengths and their relevant stopping potential, we can write (fromthe above relation) as follows:
‌

h(3×108)

0.3×10−6

=φ+(1.6×10−19)×(2)‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)
‌

h(3×108)

0.4×10−6

=φ+(1.6×10−19)×(1)‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)
‌

h(3×108)

0.5×10−6

=φ+(1.6×10−19)×(0.4)‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(3)
Now, Eq. (1)- Eq. (2) gives
h=‌

1.2×10−6

3×108

(1.6×10−19)=6.4×10−34J⋅s