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## NEET & AIIMS 11th PCB Chemistry structure of Atoms Demo Videos

Hello students, we are studying the chapter structure of atoms. So, in this module I will continue what we were discussing that is the radial probability curves. So, earlier we discussed how to calculate the total probability, right. So, for which we took a shell at a distance r which is thickness dr and we can calculate the volume of it which is 4 pi r square into dr. And probability of finding an electron in that small thickness dr it can be defined as dP which is size square into dv, where size square is the probability density function and which can be written as 4 pi r square size square into dr. So, the total probability we learned that I can integrate that from zero to P dP and it comes to be R1, R2 size square 4 pi r square dr, right, so this is what we learnt. And here that Pr that whatever that has been defined in that integral part is what we call it as a radial probability distribution function. So, therefore to calculate the total probability it is integral R1 to R2 Pr into dr, where Pr is the probability distribution function, right. This is what we learned earlier. Continuing this now we will try to draw the total probability versus the r. So, let us see that first.

So, if at all I draw the total probability versus r here, that P of r is a probability distribution function versus the r. Now, at nucleus, r is equal to zero, so Pr will be zero. So, the graph will always start from the origin. As r square increases size square is actually decreasing. So, there are two functions one is the r square part and the other one is the size square part. Then what would happen, the graph looks something like this for 1s orbital. What is it suggesting that initially size square is decreasing but the r square is increasing but as r is increasing size square is decreasing exponentially. So, initially it will increase because of the r square and then it decreases because of the size square. So, r square is contributing first, in the first half of the graph and the size square is contributing in the second half of the graph and eventually it tends to become zero.

Now, here you could see it is reaching at a maximum value and that maximum value is what we call it as radius of maximum probability and that is found to be 0.529 Angstrom which is exactly the same what Bohr’s have calculated. And let me let me assign it as a0. Now, we know that according to the Bohr’s r is equal to 0.529 n square by z, so, if z is constant for every value of n, you should be getting higher value of the radius, right. So, which I can write it as a0 into n square by z. So, from the graph it is evident that the probability of finding the electrons at nucleus and at infinite distance from the nucleus is always zero.

Now, let me draw a similar graph for 2s. For 2s if at all you see I, plotting probability distribution function versus r. Now, this is the graph you will get. Now, here what I would see that, is that we are getting a radial node here where the probability of finding electron is zero. Now, this is the maximum distance or the peak where we have the highest probability and this distance amazingly matches to 4a0 which is again suggesting from the Bohr’s theory that radius will be a0 into n square. So, it is a second orbit, so, n square gives us 4a0. So, radius of maximum probability 1s will always be less than 2s, so that is what we will get. That means electron in 2s is always found at the larger distance from the nucleus compared to that of 1s orbital.

Now, if at all I keep drawing this for different orbitals, what do I get? So, total probability, probability distribution curve versus r, if it is for 2s, this is what we found and it has one radial node, with maximum, radius of maximum probability at a distance. Similarly, if I draw it for 2P, I will get no nodes here; this is how we will get with certain maximum distance. Similarly, for 3P, you would see it is quite similar to that of 2s. Similarly, if at all I keep drawing it for, again with the radial node here and if I draw it for 3d, again you would see it is quite matching to the node of 1s and 2P. Similarly, if at all I draw it for 4d, you would see the graph varies quite similar to 2s and 3P with one radial node. This gives us as a pattern. So, here you would see there is one radial node forming so this gives us a pattern that the total number of radial nodes will be equal to n minus l minus 1. So, therefore to calculate the number of radial nodes for any particular orbital we can use this formula. That is n minus l minus 1. So, from there we can get what is the total number of radial nodes.

As we go for the next module, we will see about the plainer nodes or the angular nodes.

Thank you.

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2017-04-18T04:55:21+00:00 Categories: XI-NEET & AIIMS||0 Comments
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