Question

Light of wavelength \( 400\, \text{nm} \) falls on a metal with work function \( \phi = 2.0\, \text{eV} \). If the intensity of the light is doubled, what happens to the maximum kinetic energy of the emitted photoelectrons?

• A. It doubles
• B. It becomes zero
• C. It increases by a factor of \( \sqrt{2} \)
• D. It remains the same

Hint:

\textbf{Tip:} In photoelectric effect, intensity affects the number of emitted electrons, but not their maximum kinetic energy.

Answer 1

The Correct Option is D

The equation for the maximum kinetic energy of emitted photoelectrons in the photoelectric effect is:

\[{K_{\text{max}}}=hu-\phi\]

Where:

• \(h\) is Planck's constant (\(6.626 \times 10^{-34}\,\text{Js}\))
• \(u\) is the frequency of the incident light
• \(\phi\) is the work function of the metal

The frequency \(u\) relates to the wavelength \(\lambda\) via:

\[u=\dfrac{c}{\lambda}\]

Given \(\lambda=400\,\text{nm}=400\times10^{-9}\,\text{m}\) and the speed of light \(c=3\times10^8\,\text{m/s}\), the frequency \(u\) is calculated as:

\[u=\dfrac{3\times10^8}{400\times10^{-9}}=7.5\times10^{14}\,\text{Hz}\]

The energy of the photons is calculated as:

\[E_{\text{photon}}=hu\]

Substituting the values yields:

\[E_{\text{photon}}=6.626\times10^{-34}\times7.5\times10^{14}\approx4.97\times10^{-19}\,\text{J}\]

Converting \(E_{\text{photon}}\) to electronvolts (1 eV = \(1.602\times10^{-19}\,\text{J}\)):

\[\approx \dfrac{4.97\times10^{-19}}{1.602\times10^{-19}}\approx3.1\,\text{eV}\]

The maximum kinetic energy \({K_{\text{max}}}\) is calculated as:

\[{K_{\text{max}}}=3.1-\phi=3.1-2.0=1.1\,\text{eV}\]

An increase in light intensity doubles the number of photons striking the metal per second but does not alter the energy of individual photons or the light's frequency. Consequently, the maximum kinetic energy of the emitted photoelectrons remains constant.

The correct answer is:

• It remains the same.