The radial wavefunction, R(r), for the 2s orbital of a...

The radial wavefunction, R(r), for the 2s orbital of a hydrogenic ion/atom is: R(r) = C (Z/a0)^(3/2) (2 - ?)e^(-?/2) where C is a constant, Z is the atomic number, ? = 2Zr/(na0), r is the distance from the nucleus, n denotes the principal quantum number and a0 = 5.29 x 10^-11 m. (a) Determine the value(s) of r corresponding to the radial node(s) for He+. (b) Write the Radial Distribution Function for the 2s orbital. (c) Sketch the 2s radial wavefunction and the corresponding Radial Distribution Function. (d) Calculate the location(s) of the radial node(s) for Li2+ and the hydrogen atom. (e) Without deriving the formula, explain how the position with the highest electron density can be calculated for the 2s orbital.

Transcript

00:01 Students, welcome here in this question.

00:01 The radial wave function, rr, is given for 2 -as orbitals of hydrogenic ion.

00:23 So how we can write is, rr is equal to c, z divided by a -0, 4 power 3 divided by 2, 2 minus, row exponential power minus row divided by 2.

00:41 So if you see here, c is the constant.

00:48 So c is constant.

00:52 And z is atomic number.

01:00 And row is equal to 2 z are divided by n .a.

01:07 Not okay are is the distance distance from the nucleus and principal quantum number is n is equal to principal quantum number and a not is given value so a not is equal to 5 .29 multiplied by 10 per minus 11 meters.

01:55 So here some questions are given regarding this information.

02:01 So first one is determine the value of r corresponding to the radial nodes of helium plus.

02:12 So if you see here, how can you calculate r value? so here, z is equal to so r square, r is equal to 0 at radial node because it is s -arbital so there is a zero is a value and r square r is equal to 1 divided by 16 2 pi 2 divided by a 0 multiplied 3 over 2 multiplied by 2 minus 4 r divided by 2a0 exponential power minus 2r divided by 2a0 is a value so if you see here the radial node value now two possibilities are present here so one is e whole power 4r divided by 2 .2 multiplied by 5 .29 multiplied by 5 .29 by 10 power minus 11 is equal to 0 and the second one is you see the second one the same 2 minus r so second one is equal to 2 minus 4 r is divided by 2 multiplied by 5 .29 multiplied by 10 power minus 11 is equal to 0 so these two are then here if here then r is equal to infinity where it is which is extremely extreme at point we will see the r value so r is equal to 2 multiplied by 2 multiplied by 5 .29 multiplied by 10 power minus 11 divided by 4 then it is equal to 5 .29 multiplied by 10 power minus 11 meters is the r value radius we find out the radius here now radial distribution function for the two s orbital we have to find out so radial distribution function for the two s orbital so is equal to 4 pi r square r square r is a value so we have r square r value so 4 pi r square r square and this one is 1 divided by 4 to root 2 pi and this one is z divided by a 0 and 3 divided by 2 minus 2 z are divided by 2 a 0 exponential power 2 z r divided by 4 a not and so here n is equal to 2 we know that okay so for 2 is given here this one is whole square okay now the value is 4 pi r square so this pi and this pi will get cancelled so now it is 4 pi r square 1 divided by 32 and z divided by a 0 4 x divided by a 0 4 3 power 2 because here there is a 2 is present so it will be cancelled so only 2 will be remained here.

06:10 Now the value is equal to 2 minus z r a0 2 is given here and e exponential power minus z are divided by 4a0.

06:31 So 4 is also cancelled here to do 0 and this one is so here it is also cancelled.

06:38 Now the value will be we will see here the value is equal to r squared by a z divided by a not whole cube 2 minus z divided by a not whole square exponential power z r divided by a not okay now this is a radial function for two s orbitals now the second third one is given here, they are sketched the 2s orbital wave function and the corresponding radial distribution function...

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