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 photon is 4.7 eV
(b) the energy of the incident photon is 2.3 eV
(c) if higher frequency photon be used, the
photoelectric current will rise
(d) when the energy of photon is 3.5 eV, the
photoelectric current will be maximum

Answer 1

Explanation:

Given,

Work function, W

o

=3.5eV

Potential, V=E

k

=1.2eV

According to Einstein’s photoelectric equation,

hν=W

o

+

2

1

mv

2

hν=3.5+1.2

hν=4.7eV

Answer 2

Given:

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.

To find:

Correct statement among the options.

Calculation:

Since , the electrons can be stopped by a potential of -1.2 V , hence the kinetic energy of the electrons is 1.2 eV.

Now , applying Einstein's Photoelectric Effect Equation:

$\therefore \: \rm{KE = energy - W_{0}}$

$= > \: \rm{ 1.2= energy - 3.5}$

$= > \: \rm{ energy = 3.5 + 1.2}$

$= > \: \rm{ energy = 4.7 \: eV }$

So, final answer is:

$\boxed{ \: \bf{ energy \: of \: incident \: photons = 4.7 \: eV }}$

• Always remember that energy of Incident photons will be greater than that of work function for that metal.