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A beam of light consists of 4 wavelength $\;4000A^{0},4800A^{0},6000A^{0}\; and\;7000A^{0}\;,$ each of intensity $\;1.5\times10^{-3}Wm^{-2}\;.$The beam falls normally on an area $\;10^{-4} m^2\;$ of a clean metallic surface of work function 1.9eV . Assuming no loss of light energy the no.of photoelectrons liberated per second is


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Answer : (d) $\;1.12\times10^{12}$
Explanation :
Therefore , light of wavelength $\;4000A^{0}\;,4800A^{0}\;and\;6000A^{0}\;$ can only emit photoelectrons .
Therefore , Number of photoelectrons emitted per second = No . of photons incident per second .
answered Feb 22, 2014 by yamini.v

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