Question

The work function of Cesium is 2.27eV. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of 600nm wavelength is
A.)0.5 V
B.)-0.2 V
C.)-0.5 V
D.)0.2 V
E.)None of the above

Answer 1

Hint: We are going to find the energy of the light(photon) with a wavelength of 600nm. Then we will see what is the voltage that stops the emission of the electrons when this light falls on the cesium cathode.

Complete solution Step-by-Step:
The energy of the photon with a wavelength lambda is given by
$E = \dfrac{hc}{\lambda}$
Where h is the planck's constant
c is the speed of light
$\lambda$ is the wavelength

We are given

$\lambda$= 600nm.

We know that

$c = 3\times 10^8 m/s$
$h = 4.1357 \times 10^{-15} eV$
$E = \dfrac{4.1357 \times 10^{-15}\times 3\times 10^8 C }{600\times 10^{-9}}$
$E = \dfrac{4.1357\times 3}{6}$
E = 2.06785 eV

E is less than the work function of cesium So the electrons are not emitted. As the electrons are not emitted there is no point in discussing cutoff voltage to stop electron emission as electrons are not emitted.

Additional information :
The Photoelectric effect was proposed by Einstein and in simple terms we can explain the energy of the emitted electrons from the metal plate when a beam of photons is incident on the metal plate. The minimum energy required to eject an electron out of the metal is called the work function of the metal surface and the energy of the emitted electron is the difference between the incident photon energy and the work function.

Note: We have to be very careful with such types of questions. By applying the formula directly leads to a wrong answer. We need to first check if the condition is met or not (For example in this question whether electrons are emitted or not). We can also solve this by taking h value in J.s and then converting the value to eV.