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:

Options

A. 76.5 sec
B. 57.6 sec
C. 5.76 sec
D. 5.0 sec

Answer

B

Explanation

Intensity at the location of the ptassium surface is given by
`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`