Question 83: Dual Nature of Matter and Radiation

Question

UV light of $$4.13 \mathrm{~eV}$$ is incident on a photosensitive metal surface having work function $$3.13 \mathrm{~eV}$$. The maximum kinetic energy of ejected photoelectrons will be:

Accepted Answer

To find the maximum kinetic energy of the ejected photoelectrons, we'll use the photoelectric effect equation:
$$KE_{	ext{max}} = h
u - \phi$$
where $$KE_{	ext{max}}$$ is the maximum kinetic energy of the ejected electrons, $$h$$ is Planck's constant, $$
u$$ is the frequency of the incident light, and $$\phi$$ is the work function of the metal.
However, in this problem, we are given the energy of the UV light in electronvolts (eV) directly, which simplifies the problem. The energy of the UV light in electronvolts also represents the energy of the photons ($$h
u$$) hitting the metal surface. Thus, we can calculate the maximum kinetic energy of the ejected photoelectrons using the given energies directly:
$$KE_{	ext{max}} = E_{	ext{photon}} - \phi$$
Substituting the given values:
$$KE_{	ext{max}} = 4.13 \, 	ext{eV} - 3.13 \, 	ext{eV} = 1 \, 	ext{eV}$$
Therefore, the maximum kinetic energy of the ejected photoelectrons will be $$1 \, 	ext{eV}$$. The correct answer is
Option B: 1 eV
.

Answer

4.13 eV

Answer

1 eV

Answer

7.26 eV

Answer

3.13 eV

Dual Nature of Matter and Radiation

UV light of $4.13 \mathrm{~eV} is incident on a photosensitive metal surface having work function 3.13 \mathrm{~eV}$. The maximum kinetic energy of ejected photoelectrons will be:

Choose the correct answer: