Question 83: Dual Nature of Matter and Radiation

Question

The threshold frequency of a metal with work function $6.63 \mathrm{~eV}$ is :

Accepted Answer

The threshold frequency, $ 
u_0 $, corresponds to the minimum frequency of light required to eject electrons from the surface of a metal, a phenomenon known as the photoelectric effect. The work function, represented by $ \phi $, is the minimum energy needed to remove an electron from the surface of the metal.
The energy of a photon is given by the equation $ E = h
u $, where:
$ E $ is the energy of the photon,
$ h $ is Planck's constant ($ h = 6.626 	imes 10^{-34} \, 	ext{J} \cdot 	ext{s} $), and
$ 
u $ is the frequency of the photon.
To find the threshold frequency for a metal with a work function of $ 6.63 \, 	ext{eV} $, we must first express the work function in joules (since Planck's constant is in joules per second). To convert electron volts to joules, use the conversion factor $ 1 \, 	ext{eV} = 1.602 	imes 10^{-19} \, 	ext{J} $:
$ \phi = 6.63 \, 	ext{eV} 	imes 1.602 	imes 10^{-19} \, \frac{	ext{J}}{	ext{eV}} = 1.061 	imes 10^{-18} \, 	ext{J} $
The energy of the photon at the threshold frequency is equal to the work function:
$ h
u_0 = \phi $
Solve for $ 
u_0 $:
$ 
u_0 = \frac{\phi}{h} = \frac{1.061 	imes 10^{-18} \, 	ext{J}}{6.626 	imes 10^{-34} \, 	ext{J} \cdot 	ext{s}} = 1.6 	imes 10^{15} \, 	ext{Hz} $
Thus, the correct answer is:
Option C
$ 1.6 	imes 10^{15} \, 	ext{Hz} $

Answer

$16 	imes 10^{15} \mathrm{~Hz}$

Answer

$16 	imes 10^{12} \mathrm{~Hz}$

Answer

$1.6 	imes 10^{15} \mathrm{~Hz}$

Answer

$1.6 	imes 10^{12} \mathrm{~Hz}$

Dual Nature of Matter and Radiation

The threshold frequency of a metal with work function 6.63 \mathrm{~eV} is :

Choose the correct answer: