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Ultraviolet light of wavelength 66.26 nm and intensity `2 W//m^(2)` falls on potassium surface by which photoelectrons are ejected out. If only 0.1% of the incident photons produce photoelectrons, and surface area of metal surface is `4 m^(2)`, how many electrons are emitted per second?A. `2.67 xx 10^(15)`B. `3 xx 10^(15)`C. `3.33 xx 10^(7)`D. `4.17 xx 10^(16)` |
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Answer» Correct Answer - A Number of photons falling on metal surface, `n_(p)=("intensity"xx"area")/("energy per quanta")` `=((2Js^(-1)m)xx(4m^(2)))/({{(6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/((66.26xx10^(-9)m)}})` `=2.67xx10^(18)" per sec"` From equation, `n_(e)=0.1%" of "n_(p)` `=(0.1)/(100)xx2.67xx10^(18)=2.67xx10^(15)" per sec"` |
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