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 maximum kinetic energy of emitted photoelectrons in the photoelectric effect is determined by the equation: 

\[{K_{\text{max}}}=h\nu-\phi\]

Where:

• \(h =\) Planck's constant (\(6.626 \times 10^{-34}\,\text{Js}\))
• \(\nu =\) Frequency of the incident light
• \(\phi =\) Work function of the metal

The frequency \(\nu\) can be related to the wavelength \(\lambda\) using:

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

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

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

The energy of the photons is obtained by:

\[E_{\text{photon}}=h\nu\]

Substituting the values:

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

Convert \(E_{\text{photon}}\) 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:

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

Doubling the intensity of light increases the number of photons hitting the metal per second, but it does not change the energy per photon or the frequency of light. Therefore, the maximum kinetic energy of the emitted photoelectrons remains unchanged.

The correct answer is:

• It remains the same.