1.

Light of wavelength 5000Åand intensity 39.8Wm^(-1) is incident on a metal surface. If only 1% photons of incident light emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be nearly :

Answer»

`10^(18)`
`10^(20)`
`10^(22)`
`10^(24)`

Solution :`E=nh RARR E=(nhc)/(lambda)`
`rArr n=(E lambda)/(hc)`
But `n=(n)/(100)=(E lambda)/(100 hc)=10^(18)`


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