00:01
So here's a question if a photon of wavelength 0 .04250 nanometer.
00:09
So wavelength lambda, lambda is equal to 0 .04250 nanometer.
00:21
Okay.
00:22
It strikes a free electron and is scattered at angle of 35 degree.
00:26
So theta is 35 degree.
00:29
From its original direction, find change in wavelength.
00:33
So first part change in wavelength that is you have to calculate delta lambda.
00:40
So delta lambda can be given by h upon mc into 1 minus cos theta.
00:47
Here mass will be mass of electron.
00:52
C is speed of light, h is a high as a constant.
00:54
Let's put the value 6 .626 into 10 to the power minus 34.
01:00
Fine into one minus cost of 35 degree all right whole divided by mass of electron that is 9 .11 into 10 to the power minus 31 into 3 into 10 to the power 8 if we calculate its value it will comes up to be 4 .384 into 10 to the power minus 13 meter or we can say 0 .00 4384 nanometer fine so this is our delta lambda all right now let's come into the second part second part part b what it is saying it is saying the wave to find the wavelength of the scattered light okay so change in wavelength will be delta lambda will be the new lambda minus the old lambda old wavelength before the before striking before the photon has striked so it will be the new lambda or the changed lambda changed wavelength will be delta lambda plus lambda so delta lambda is we have got how much 4 .384 into 10 to the power minus 13 or better we do in nanometers.
02:45
Alright, so it is how much? 0 .004384 plus 0 .04 plus 0 .04250.
02:58
So the change lambda or the new lambda is 0 .04293 nanometers.
03:10
It can also be written as 4 .293 into 10 to the power minus 11 meters.
03:18
Fine...