A beam of light has two wavelengths of 4972,m | vedclass

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(B) Given: $\lambda_1 = 4972\,\mathring{A}$,$\lambda_2 = 6216\,\mathring{A}$,Total Intensity $I = 3.6 	imes 10^{-3}\,	ext{W/m}^2$,Area $A = 1\,	ext

Vedclass · Accepted Answer

$9 	imes 10^{11}$

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(B) Given: $\lambda_1 = 4972\,\mathring{A}$,$\lambda_2 = 6216\,\mathring{A}$,Total Intensity $I = 3.6 	imes 10^{-3}\,	ext{W/m}^2$,Area $A = 1\,	ext{cm}^2 = 10^{-4}\,	ext{m}^2$,Work function $\phi = 2.3\,	ext{eV}$.Intensity per wavelength $I' = I/2 = 1.8 	imes 10^{-3}\,	ext{W/m}^2$.Energy of photons for each wavelength:$E_1 = \frac{hc}{\lambda_1} = \frac{12400}{4972} \approx 2.49\,	ext{eV} > 2.3\,	ext{eV}$ (Capable of emission).$E_2 = \frac{hc}{\lambda_2} = \frac{12400}{6216} \approx 1.99\,	ext{eV} Only photons of $\lambda_1$ contribute to photoelectric emission.Power incident from $\lambda_1$ is $P = I' 	imes A = 1.8 	imes 10^{-3} 	imes 10^{-4} = 1.8 	imes 10^{-7}\,	ext{W}$.Number of photons per second $n = \frac{P}{E_1} = \frac{1.8 	imes 10^{-7}}{2.49 	imes 1.6 	imes 10^{-19}} \approx 4.5 	imes 10^{11}\,	ext{photons/s}$.Since each capable photon ejects one electron,the number of photoelectrons liberated in $2\,	ext{s}$ is $N = n 	imes 2 = 4.5 	imes 10^{11} 	imes 2 = 9 	imes 10^{11}$.

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