When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:

1.0 eV

1.3 eV

1.5 eV

0.65 eV

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is: Correct Answer 1.0 eV

CONCEPT:

When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect.
Threshold frequency: The minimum frequency required to emit an electron from the metal surface is called Threshold frequency.
- It is denoted by ν₀.
- The kinetic energy of the photoelectrons was proportional to the light frequency.
The total energy of the incoming photon E_photon, must be equal to the kinetic energy of the ejected electron, KE_electron,plus the energy required to eject the electron from a particular metal is also called Metal's work function, It is represented by the symbol ϕ (in units of J)

E_photon = KE_electron + ϕ

E_photon = hv = KE_electron+ ϕ

Rearranging this equation in terms of the electrons kinetic energy, we get:

KE_electron = hv - ϕ

EXPLANATION:

For photoelectric equation,

(KE)_min = hv - ϕ_o

For the first condition,

0.5 = E - ϕ_o ----------(i)

For the second condition,

0.8 = 1.2 E -ϕ_o --------(ii)

from equations (i) and (ii)

0.5 = E - ϕ_o and

0.8 = 1.2 E - ϕ_o

-0.3 = -0.2E

E =(0.3)/(0.2) = 1.5 eV

From expression (i)

0.5 = 1.5 - ϕ_o

=> ϕ_o = 0.5 - 0.5

= 1 eV

option 1 is answer.