A source of monochromatic radiation of wavele | vedclass

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Question

(A) Total energy provided by the source per second $= \frac{1000 \, J}{10 \, s} = 100 \, J/s$.Energy of one photon $E = \frac{hc}{\lambda} = \frac{6.6

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Total energy provided by the source per second $= \frac{1000 \, J}{10 \, s} = 100 \, J/s$.Energy of one photon $E = \frac{hc}{\lambda} = \frac{6.626 	imes 10^{-34} \, Js 	imes 3 	imes 10^8 \, m/s}{400 	imes 10^{-9} \, m} = 4.9695 	imes 10^{-19} \, J$.Number of photons incident per second $= \frac{	ext{Total energy per second}}{	ext{Energy of one photon}} = \frac{100}{4.9695 	imes 10^{-19}} \approx 2.012 	imes 10^{20}$.Since one photon ejects one electron,the number of electrons ejected per second is $2.012 	imes 10^{20}$.Comparing this with $x 	imes 10^{20}$,we get $x \approx 2$.

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