Question
In photoelectric effect, the work function of a metal is $$3.5\,eV.$$ The emitted electrons can be stopped by applying a potential of $$- 1.2\,V.$$ Then,
A.
the energy of the incident photons is $$4.7\,eV$$
B.
the energy of the incident photons is $$2.3\,eV$$
C.
if higher frequency photons be used, the photoelectric current will rise
D.
when the energy of photons is $$3.5\,eV,$$ the photoelectric current will be maximum
Answer :
the energy of the incident photons is $$4.7\,eV$$
Solution :
When a photon of light of frequency $$\nu $$ is incident on a photosensitive metal surface. Then, according to Einstein's photoelectric equation
$$h\nu = {W_0} + \frac{1}{2}m{\nu ^2}$$
where, $${W_0}$$ is work function of the metal and $$\frac{1}{2}m{v^2}$$ is $$KE$$ of the photoelectron.
Given, $${W_0} = 3.5\,eV,\,KE = 1.2\,eV$$
$$\therefore h\nu = 3.5 + 1.2 = 4.7\,eV$$
NOTE
Stopping potential always gives maximum kinetic energy of ejected electrons.