Respuesta :
The question contains a typo error and with the help of the text editor I was able to put them down in the correct form. So, here is it.
In a one electron system, the probability of finding the electron within a shell of thickness [tex]\delta r[/tex] at a radius of r from the nucleus is given by the radial distribution function P (r) = r²R²(r)
An electron in a 1s hydrogen orbital has the radial wave function R(r) given by:
[tex]R(r) = 2|\frac{1}{a_o}|^{3/2}} \ e ^{r/a_o}[/tex] where [tex]a__0[/tex] is the Bohr radius (52.9 pm)
Calculate the probability of finding the electron in a sphere of radius 1.1 [tex]a__0[/tex]centered at the nucleus.
Answer:
Probability (P) ≅ 37.73 %
Step-by-step explanation:
[tex]\\ \\ \\Probability \ (P) = \int\limits^b_0 \ P({r}) \, \ dr \\ \\ \\ where \ b= 1.1a__0}} \\ \\ = \int\limits^b_0 r^2 \ 4 (\frac{1}{a__0}})^3 \ e^{-2r/a__0}} \ dr \\ \\ \\ = \frac{4}{a^3__0}}} \ \int\limits^b_r r^2 \ e^{-2r/a__0}} \ dr \\ \\ \\ = \frac{4}{a^3__0}}} [(r^2 \frac{e^{-2r/a_0}}{(-2/a_0_})}]^b___0}}} \ - \ \int\limits^b_0 2r \frac{e^{-2r/a}}{-2/a_0} \ dr][/tex]
[tex]= \frac{4}{a__0}^3}[(\frac{a_0b^2e^{-2b/a_0}}{(-2)})+a_0\int\limits^b_0 \ r \ e^{-2r/a} \ dr] -------- equation (1)[/tex]
However:
[tex]\int\limits^b_0 \ r \ e^{-2r/a} \ dr \\ \\ \\ = [r \ \frac{e^{-2r/a}}{(-2/a)}]^b__0}}- \int\limits^b_0 \frac{e^{-2r/a}}{(-2/a)}} \\ \\ \\ = \frac{ab \ e^{-2b/a}}{(-2)} + \frac{a}{2} \ \int\limits^b_a \ e^{-2r/a} \ dr[/tex]
[tex]= \frac{ab \ e^{-2b/a}}{(-2)} + \frac{a}{2} \frac {[e^{-2r/a}]^b_0}{(-2/a)}[/tex]
[tex]\int\limits^b_0 \ r \ e^{-2r/a} \ dr = \frac{ab \ e^{-2b/a}}{(-2)} - \frac{a^2}{4}[e^{-2b/a}-1][/tex]
Replacing the value of [tex]\int\limits^b_0 \ r \ e^{-2r/a} \ dr[/tex] into equation (1); we have:
[tex]Probability (P) = \frac{4}{a__0}^3}[(\frac{a_0b^2e^{-2b/a_0}}{(-2)})+ \frac{ab \ e^{-2b/a}}{(-2)} - \frac{a^2}{4}[e^{-2b/a}-1]][/tex]
Substituting b = 1.1 [tex]a__0[/tex] ; we have:
[tex]Probability (P) = \frac{4}{a^3}[(\frac{a^3(1.1)^2e^{-2(1.1)}}{(-2)})+ {a^2_3(1.1) \ e^{-2*1.1}} - \frac{a^3}{4}[e^{-2(1.1)}-1]][/tex]
Probability (P) = 4 (-0.067036 - 0.06094 + 0.222299)
Probability (P) = 0.377292
Probability (P) ≅ 37.73 %
Answer:
Step-by-step explanation:
Solution:-
- The conclusions of Bohr's study have gave us hints regarding the probability of finding an electron ( e- ) in a 3 dimensional space of a nucleus.
- In Bohr's model the the 3-dimensional was considered as a spherical shell with thickness ( t = δr ) . Where ( r ) is the absolute radius of the density of electrons ( e - ) found in the vicinity of a nucleus.
- Bohr performed several experiments to determine what is the probability of finding of finding a single electron ( e- ) in a atom around its nucleus.
- He found that the probability P of finding an electron is a function of radial distance ( r )^2 - square of its distance from nucleus and the atom's wave-function R ( r ). The probability of the distribution is given as:
[tex]P (r ) = r^2*R^2 ( r )[/tex]
- Where R ( r ) is the wave-function specific for an atom. Here we will investigate an Hydrogen atom which has an orbital configuration = 1s orbitals.
[tex]R ( r ) = 2*(\frac{1}{a_o} )^(^\frac{3}{2}^) * e^(^-^\frac{r}{a_o}^)[/tex]
Where, [tex]a_o = 52.9 pm[/tex] , is the Bohr's radius.
- We will determine the probability P ( r ) of finding that electron in a hydrogen atom at a radial distance r = 1.1a_o.
- Determine the P ( R ) by performing an integral from the center of spherical shell i.e nucleus r = 0 to r = 1.1a_o:
[tex]P ( r ) = \int [ 2*(\frac{1}{a_o} )^(^\frac{3}{2}^) * e^(^-^\frac{r}{a_o}^) ] ^2*r^2 dr\\\\P ( r ) = \int [ 4*(\frac{1}{a_o} )^(^3^) * e^(^-^\frac{2r}{a_o}^) ] *r^2 . dr\\\\P ( r ) = 4*(\frac{1}{a_o} )^(^3^) \int [ e^(^-^\frac{2r}{a_o}^) . r^2 ] . dr[/tex]
- Perform integration by parts:
[tex]P ( r ) = 4*(\frac{1}{a_o} )^(^3^) * ( [ \frac{e^(^-^\frac{2r}{a_o}^) }{\frac{-2}{a_o} } ]*r^2 - \int [ e^(^-^\frac{2r}{a_o}^) . 2r ] . dr)[/tex]
