Question

The work function of aluminum is 4.2eV. Light of wavelength $2000\mathring{A}$ is incident on it. If the intensity of incident light is doubled, then the value of maximum energy will become:

• A. double
• B. four times
• C. one fourth
• D. unchanged

Answer 1

The correct option is D unchanged

The intensity of incident light on it is proportional to the number of photons incident on it. For each photon, one electron is ejected. The maximum energy of the electron depends only on the frequency of the incident radiation and not on the intensity of the incident radiation. Hence, maximum energy will be unchanged.
Option D is correct.