Question 206: Dual Nature of Matter and Radiation

Question

A metal plate of area 1 $$ 	imes $$ 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons.
The number of emitted photoelectrons per second and their maximum energy, respectively, will be

Accepted Answer

Given that the area of the metal plate $$A = 1 	imes 10^{-4} m^2$$, and the intensity of the radiation
$$I = 16 mW/m^2 = 16 	imes 10^{-3} W/m^2$$.
The power $$P$$ falling on the metal plate is given by the product of the intensity of the radiation and the area:
$$P = IA = (16 	imes 10^{-3} W/m^2)(1 	imes 10^{-4} m^2) = 1.6 	imes 10^{-6} W$$.
The energy of each incident photon is given as 10 eV. Converting this to joules (since 1 eV = $$1.6 	imes 10^{-19} J$$):
$$E_{	ext{photon}} = 10 eV = 10 	imes 1.6 	imes 10^{-19} J = 1.6 	imes 10^{-18} J$$.
The number of incident photons per second $$n_{	ext{photon}}$$ is then given by the total power divided by the energy per photon:
$$n_{	ext{photon}} = P / E_{	ext{photon}} = 1.6 	imes 10^{-6} W / 1.6 	imes 10^{-18} J = 10^{12} 	ext{ photons/s}$$.
Given that only 10% of these photons actually produce photoelectrons, the number of photoelectrons produced per second $$n_{	ext{electron}}$$ is:
$$n_{	ext{electron}} = 0.1 	imes n_{	ext{photon}} = 0.1 	imes 10^{12} = 10^{11} 	ext{ electrons/s}$$.
The work function of the metal is given as 5 eV, which is the energy required to remove an electron from the metal. Therefore, the maximum energy of the photoelectrons $$E_{	ext{max}}$$ is given by the energy of the incident photons minus the work function:
$$E_{	ext{max}} = E_{	ext{photon}} - 	ext{Work function} = 10 eV - 5 eV = 5 eV$$.

Answer

1014 and 10 eV

Answer

1012 and 5 eV

Answer

1011 and 5 eV

Answer

1010 and 5 eV

Dual Nature of Matter and Radiation

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