JEE MAIN - Physics (2024 - 9th April Evening Shift - No. 9)
Explanation
To find the maximum kinetic energy of the ejected photoelectrons, we'll use the photoelectric effect equation:
$$KE_{\text{max}} = h\nu - \phi$$where $$KE_{\text{max}}$$ is the maximum kinetic energy of the ejected electrons, $$h$$ is Planck's constant, $$\nu$$ 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 ($$h\nu$$) hitting the metal surface. Thus, we can calculate the maximum kinetic energy of the ejected photoelectrons using the given energies directly:
$$KE_{\text{max}} = E_{\text{photon}} - \phi$$Substituting the given values:
$$KE_{\text{max}} = 4.13 \, \text{eV} - 3.13 \, \text{eV} = 1 \, \text{eV}$$Therefore, the maximum kinetic energy of the ejected photoelectrons will be $$1 \, \text{eV}$$. The correct answer is Option B: 1 eV.
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