Question 104: 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: