30These are doubts. Can someone help please?
1. A small 10W source of ultraviolet light of wavelength 99nm is held at a distance 0.1m from a metal surface. The radius of an atom of the metal is approximately 0.05nm. Find
a) the average number of photons striking an atom per second.
b) the number of photoelectrons emitted per unit area per second if the efficiency of liberation of photoelectrons is 1%.
2. A radioactive decay counter is switched on at t=0. A β- active sample is present near the counter. The counter registers the number of β- particles emitted by the sample. The counter registers 1 x 105 β- particles at t=36s and 1.11 x 105 β- particles at t=108s. Find t1/2 of this sample.
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13 Answers
1listen ashish the key idea here in the first question is wid help of solid angle considering the waves emitting from the source to be of equal intensity we can say that the radius of the circle on which the light is falling will be .1nm from there area can be found . i hv prceed this mch
1i think the ans for 1 (a) is 1.245
and how to do (b) since wrk fnctn is not given
12. t1/2 = 478.11 secs
i used chem kinetics here,its a 1st order rctn....u have to find λ i guess
30@Debosmit - Can you post your working. I was getting a rather strange answer.
1i think in (b) the total no of atom present in the area from there 1% will emit the photo electrons .
what do u think.........
1since the βsample emits particles linearly....the counter will record the no. of particles linearly as well....
dn/dt = λn
hence, ln(n/n0) = λt + c
now we know the data so put them....2 eqtns,subtraction will do it....
30The answers given are :
1. (a) - 5/16
(b) - 102080pi
2. t1/2=10.8s
1if the answr is 10.8secs and t1/2 is cnstnt since its a 1st order rctn the decay shud be cmplete in 21.6 secs..then the data is diffclt to interpret..can an expert pls validate??
30@Debosmit - The reaction won't complete in 21.6 secs. It will become half of what it was at 10.6 secs.
1@ashish:yes,i was wrong....but i tried again and got the same answr....
1okay i`ve got t1/2≈10.4secs after some approximations.
not a good method..anyway
the eqtns are:[N is the initial no. of β-particles in the sample].
ln(N/N-105) = λx36
ln(N/N-1.1x105) = λx108
2eqtns,divide them,u`l get a cubic eqtn....cancel out N from both sides by tkng common since N≠0
a qdrtc eqtn....solve
30Me getting same.. I left the calculations after I reached the rather dirty equation!