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  • A beam of light has three wavelengths , <img alt="\mathrm{4972 \AA }" src="https

A beam of light has three wavelengths \mathrm{4144 \AA }, \mathrm{4972 \AA } and \mathrm{ 6216 } \AA with a total intensity of \mathrm{3.6 \times 10^{-3} Wm^{-2}} equally distributed amongst the three wavelengths. The beam falls normally on an area \mathrm{1.0\: cm^{2}} of a clean metallic surface of work function \mathrm{ 2.3 eV}. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds.

Option: 1

1.1\times10^{12}


Option: 2

0.8\times10^{12}


Option: 3

0.3\times10^{12}


Option: 4

1.4\times10^{12}


Answers (1)

best_answer

Energy of photon having wavelength \mathrm{4144 \AA,}

\mathrm{ E_1=\frac{12375}{4144} \mathrm{eV}=2.99 \mathrm{eV} }

Similarly,

\mathrm{ E_2=\frac{12375}{4972} \mathrm{eV}=2.49 \mathrm{eV} \text { and } }

\mathrm{ E_3=\frac{12375}{6216} \mathrm{eV}=1.99 \mathrm{eV} }

Since, only \mathrm{ E_1 \: and \: E_2 } are greater than the work function \mathrm{ W=2.3 \mathrm{eV} }, only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally distributed in all wavelengths. Therefore, intensity corresponding to each wavelength is

\mathrm{ \frac{3.6 \times 10^{-3}}{3}=1.2 \times 10^{-3} \mathrm{watt} / \mathrm{m}^2 }
Or energy incident per second in the given area \mathrm{ \left(A=1.0 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2\right) } is :

\mathrm{ \rho=1.2 \times 10^{-3} \times 10^{-4}=1.2 \times 10^{11} }

Let \mathrm{ n_1 } be the number of photons incident per unit time in the given area corresponding to first wavelength. Then,

\mathrm{ n_1 =\frac{\rho}{E_1}=\frac{1.2 \times 10^{-7}}{2.99 \times 1.6 \times 10^{-19}} }

\mathrm{ =3.0 \times 10^{11} }

Similarly,

\mathrm{ n_2 =\frac{\rho}{E_2}=\frac{1.2 \times 10^{-7}}{2.49 \times 1.6 \times 10^{-19}} }

\mathrm{ =3.0 \times 10^{11} }

Since, each energetically capable photon ejects, total number of photons liberated in 2 seconds

\mathrm{ =2\left(n_1+n_2\right) }

\mathrm{ =2(2.5+3.0) \times 10^{11} }

\mathrm{ =1.1 \times 10^{12} }

 
 

Posted by

Divya Prakash Singh

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