00:01
The question given is from the topic electronic configuration and the product table.
00:05
In this question it is asked that find the energy of photon which has a blue light and that is a wavelength of 450 nanometers.
00:12
So in this question we need to use the concept of energy formula finding using lambda given.
00:18
So let me first tell you the formula that we use but before that we'll do the data writing part.
00:25
So in this question we have been given the wavelength and how do we represent wavelength? yeah, we have represented with lambda.
00:33
We use lambda for wavelength and that value given is 450 nanometers and you've been asked to find the energy.
00:44
Now we've been asked to find the energy of the photon.
00:49
How do you find the energy of the photon? so energy of the photon e, energy of photon e, is calculated by the formula hc by lambda.
01:02
H c by lambda where what is h is called as planx constant which is 6 .62 into 10 to the power minus 34 joules second and the c is called as the speed of light as we know which is 3 into 10 to the power 8 meter per second.
01:19
Now if you put in the formula and the formula all the values we'll get something like this the energy would be 6 .62 into 10 to the power minus 34 into 3 into 10 to the power 8 joules second and second second gets cancelled, jule per meter...