Question 67: Dual Nature of Matter and Radiation

Question

The work function of a metal is 3 eV . The color of the visible light that is required to cause emission of photoelectrons is

Accepted Answer

To determine which color of visible light will cause the emission of photoelectrons from a metal with a work function of 3 eV, we need to consider the relationship between the photon energy, the work function, and the wavelength of light.
The maximum kinetic energy ($ 	ext{KE}_{\max} $) of the emitted photoelectrons is given by:
$ 	ext{KE}_{\max} = \frac{hc}{\lambda} - \phi $
For photoemission to occur, the energy of the incident photons must be greater than the work function ($\phi$):
$ \frac{hc}{\lambda} > \phi $
This implies:
$ \lambda < \frac{hc}{\phi} $
Substitute the values $ h = 4.135667696 	imes 10^{-15} \, 	ext{eV}\cdot	ext{s} $, $ c = 3 	imes 10^{8} \, 	ext{m/s} $, and $\phi = 3 \, 	ext{eV}$:
$ \lambda < \frac{1242}{3} \, 	ext{nm} \approx 414 \, 	ext{nm} $
Since the wavelength must be less than 414 nm to cause emission, we need light with a shorter wavelength. Among the visible colors, blue light has a wavelength in this range, making it the correct choice. Thus, blue light will cause the emission of photoelectrons with this work function.

Answer

Red

Answer

Green

Answer

Blue

Answer

Yellow

Dual Nature of Matter and Radiation

The work function of a metal is 3 eV . The color of the visible light that is required to cause emission of photoelectrons is

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