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A potassium surface is placed 75 cm away from a 100 W bulb. It is found that energy radiated by the bulb is 5% of the input power. Consider each potassium atom as a circular disc of diameter 1Å and determine the time required for each atom to absorb an amount of energy equal to its work function of 2:0 eV: |
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Answer» 76.5 sec `I=("POWER of bulb")/("Areaof sphere")=(100xx(5)/(100))/(4pixx(0*75)^(2))` `=0*707 wm^(-2)` The power of incident on each potassium atom is `P_(i)`= intensity `xx` area per atom `=0*707xx(pixx(10^(-10))^(2))/(4)=5*56xx10^(21-)W` `:. "Time" =("Energy")/("Power")=(W)/(P_(i))=(2xx1*6xx10^(-19))/(5*56xx10^(-21))` `=57*6SEC` |
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