Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

A small potassium foil is placed (perpendicular to direction of incidence of light ) a distance r (=0.5 m) from a point light source whose output power $\;P_{0}\;$ is 1W . Assuming wave nature of light . how long would it take for foil to soak up enough energy (=1.8eV) from the beam to eject on electron ? (Assume that ejected photoelectron collected its energy from a circular area of foil whose radius equals radius of potassium atom ($\;1.3\times10^{-10}m$))


Can you answer this question?

1 Answer

0 votes
Answer : (b) 17s
Explanation :
If source radiates uniformly in all direction , the intensity I of light at a distance r is given by
$I=\large\frac{P_{0}}{4 \pi r^2}=\large\frac{1 W}{4 \pi \;(0.5 m)^2}=0.32 W/m^2$
The target area A is $\;\pi\;(1.3\times10^{-10}m)^2\quad \;5.3\times10^{-20} m $
So the rate at which energy falls on target is given by $\;P=IA=(0.32)\;5.3\times10^{-20}$
If all this incoming energy is absorbed , the time required to accumulate enough energy for the electron to escape is
$t=(\large\frac{1.8eV}{1.7\times10^{-20} J/s})\;(\large\frac{1.6\times10^{-19} J}{1eV})=17s\;.$
answered Feb 22, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App