Question

The radiations of 500 nm liberate photo
electrons from the cesium surface of work
function 1.8 ev. The energy of electron is
.......... eV.
1.0.068 2.0.168 3.0.68 4.0.681​

Answer 1

Answer:

3) 0.68eV

Explanation:

We make use of Einstein's photoelectric equation to solve this question

Refer attachement for detailed solution

Hope this answer helped you ✌

Answer 2

Answer:

The energy of the electron is 0.68eV.

Explanation:

In this question, we will utilize the concept of the photoelectric effect.

The process of emission of electrons from a metal surface, when electromagnetic radiations of sufficiently high energy is incident on it, is called the photoelectric effect.

Einstein explained the photoelectric effect based on Planck's quantum theory according to which light radiation travels in the form of discrete photons called quanta . The energy of each photon is hv, where h is Planck's constant and v is the frequency of light. The relation can be written as,

E=Ф°+Kmax      (1)

E=energy of photon

Ф°=work function of metal

Kmax=maximum kinetic energy

first, let's find E,

E=hc/λ'

λ=500nm

c= 3 x $10^{8}$m/s

h=6.626 x $10^{-34}$

after substituting the values we will get,

E=4 x $10^{-19}$J

To convert it into ev we need to divide it by 1.6 x $10^{-19}$

after dividing we will get,

E=2.5eV

lets substitute the values in equation 1,

2.5=1.8+K

K=0.7ev

among the given option the closest answer will be 0.68

so, 0.68(3) will be the right answer