JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 15)
Given below are two statements:
Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.
Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.
In the light of above statements, choose the correct answer form the options given below
Both Statement I and Statement II are true
Statement I is true but statement II is false
Statement I is false but statement II is true
Both Statement I and Statement II are false
Explanation
Now let's analyze both statements:
Statement I is correct. According to the photoelectric effect, the ability to emit electrons from a metallic surface depends on the energy of the incident light. The energy of a photon is given by $$E = h\nu$$, where $$h$$ is Planck's constant and $$\nu$$ is the frequency of the light. Ultraviolet rays have higher frequencies and thus higher energies compared to microwaves and infrared rays, making them more effective for the emission of electrons from a metallic surface.
Statement II is incorrect. The maximum kinetic energy of photoelectrons is given by $$K_{max} = h\nu - h\nu_0$$, where $$\nu_0$$ is the threshold frequency. This equation shows that the maximum kinetic energy of photoelectrons is directly proportional to the frequency of the incident light (above the threshold frequency), not inversely proportional.
Based on the analysis, the correct answer is:
Statement I is true, and Statement II is false.
Statement I is correct. According to the photoelectric effect, the ability to emit electrons from a metallic surface depends on the energy of the incident light. The energy of a photon is given by $$E = h\nu$$, where $$h$$ is Planck's constant and $$\nu$$ is the frequency of the light. Ultraviolet rays have higher frequencies and thus higher energies compared to microwaves and infrared rays, making them more effective for the emission of electrons from a metallic surface.
Statement II is incorrect. The maximum kinetic energy of photoelectrons is given by $$K_{max} = h\nu - h\nu_0$$, where $$\nu_0$$ is the threshold frequency. This equation shows that the maximum kinetic energy of photoelectrons is directly proportional to the frequency of the incident light (above the threshold frequency), not inversely proportional.
Based on the analysis, the correct answer is:
Statement I is true, and Statement II is false.
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