The beam of light has three wavelengths 4144 | vedclass

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Question

(B) The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{\phi} = \frac{12420 \, eV \cdot \mathring A}{2.3 \, eV} = 5400 \, \mathrin

Vedclass · Accepted Answer

$1.075 	imes 10^{12}$

Vedclass · Answer

(B) The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{\phi} = \frac{12420 \, eV \cdot \mathring A}{2.3 \, eV} = 5400 \, \mathring A$. Only wavelengths smaller than $\lambda_0$ can eject photoelectrons. Thus,only $\lambda_1 = 4144 \, \mathring A$ and $\lambda_2 = 4972 \, \mathring A$ contribute to the photoelectric effect. The total intensity is $I = 3.6 	imes 10^{-5} \, W/m^2$. Since it is equally distributed,the intensity for each wavelength is $I_i = 1.2 	imes 10^{-5} \, W/m^2$. The area $A = 1 \, cm^2 = 10^{-4} \, m^2$. The power incident for each wavelength is $P_i = I_i 	imes A = 1.2 	imes 10^{-5} 	imes 10^{-4} = 1.2 	imes 10^{-9} \, W$. In $t = 2 \, s$,the energy incident for each wavelength is $E_i = P_i 	imes t = 1.2 	imes 10^{-9} 	imes 2 = 2.4 	imes 10^{-9} \, J$. The number of photons $n_i$ for each wavelength is $n_i = \frac{E_i \lambda_i}{hc}$. For $\lambda_1 = 4144 \, \mathring A$: $n_1 = \frac{2.4 	imes 10^{-9} 	imes 4144 	imes 10^{-10}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} \approx 0.5 	imes 10^{12}$. For $\lambda_2 = 4972 \, \mathring A$: $n_2 = \frac{2.4 	imes 10^{-9} 	imes 4972 	imes 10^{-10}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} \approx 0.6 	imes 10^{12}$. Total photoelectrons $N = n_1 + n_2 = 1.1 	imes 10^{12} \approx 1.075 	imes 10^{12}$.

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