Question 232: Dual Nature of Matter and Radiation

Question

The threshold frequency for a metallic surface corresponds to an energy of $$6.2$$ $$eV$$ and the stopping potential for a radiation incident on this surface is $$5V.$$ The incident radiation lies in

Accepted Answer

The energy of the incident radiation that causes photoelectrons to be emitted can be calculated using the stopping potential, V, via the equation
$$E = eV,$$
where e is the elementary charge.
The elementary charge, e, is approximately equal to $$1.602 	imes 10^{-19}$$ C. So the energy of the incident radiation is
$$E = 1.602 	imes 10^{-19} \, 	ext{C} 	imes 5 \, 	ext{V} = 8.01 	imes 10^{-19} \, 	ext{J}.$$
We convert this to electron volts (eV) by using the conversion factor $$1 \, 	ext{eV} = 1.602 	imes 10^{-19} \, 	ext{J}.$$ So
$$E = \frac{8.01 	imes 10^{-19} \, 	ext{J}}{1.602 	imes 10^{-19} \, 	ext{J/eV}} = 5 \, 	ext{eV}.$$
This is less than the threshold energy of $$6.2 \, 	ext{eV},$$ which means that the incident radiation does not have enough energy to overcome the work function of the metal and thus cannot cause photoelectrons to be emitted.
The regions of the electromagnetic spectrum are generally classified by energy as follows:
Infrared: Less than about $$1.24 \, 	ext{eV}$$
Visible: Between about $$1.24 \, 	ext{eV}$$ and $$3.1 \, 	ext{eV}$$
Ultraviolet: Between about $$3.1 \, 	ext{eV}$$ and $$124 \, 	ext{eV}$$
X-rays: Greater than about $$124 \, 	ext{eV}$$
Therefore, the incident radiation falls in the ultraviolet region.

Answer

ultra-violet region

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infra-red region

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visible region

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$$x$$-ray region

Dual Nature of Matter and Radiation

The threshold frequency for a metallic surface corresponds to an energy of $6.2 eV and the stopping potential for a radiation incident on this surface is 5V.$ The incident radiation lies in

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