Mathematics, 24.04.2020 04:31 janessa0804

In a one electron system, the probability of finding the electron within a shell of thickness rδr at a radius of rr from the nucleus is given by the radial distribution function, P(r)=r2R2(r)P(r)=r2R2(r) . An electron in a 1s1s hydrogen orbital has the radial wavefunction R(r)R(r) given by R(r)=2(1a0)3/2e−r/a0 R(r)=2(1a0)3/2e−r/a0 where a0a0 is the Bohr radius (52.9 pm)(52.9 pm) .

Calculate the probability of finding the electron in a sphere of radius 1.1a01.1a0 centered at the nucleus.

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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 $\delta r$ 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:

$R(r) = 2|\frac{1}{a_o}|^{3/2}} \ e ^{r/a_o}$ where a__0 is the Bohr radius (52.9 pm)

Calculate the probability of finding the electron in a sphere of radius 1.1 a__0 centered at the nucleus.

Probability (P) ≅ 37.73 %

Step-by-step explanation:

$\\ \\ \\Probability \ (P) = \int\limits^b_0 \ P({r}) \, \ dr \\ \\ \\ where \ b= 1.1a__0}} \\ \\ = \int\limits^b_0 r^2 \ 4 (\frac{1}{a__0}})^3 \ e^{-2r/a__0}} \ dr \\ \\ \\ = \frac{4}{a^3__0}}} \ \int\limits^b_r r^2 \ e^{-2r/a__0}} \ dr \\ \\ \\ = \frac{4}{a^3__0}}} [(r^2 \frac{e^{-2r/a_0}}{(-2/a_0_})}]^b___0}}} \ - \ \int\limits^b_0 2r \frac{e^{-2r/a}}{-2/a_0} \ dr]$

$= \frac{4}{a__0}^3}[(\frac{a_0b^2e^{-2b/a_0}}{(-2)})+a_0\int\limits^b_0 \ r \ e^{-2r/a} \ dr] -------- equation (1)$

However:

$\int\limits^b_0 \ r \ e^{-2r/a} \ dr \\ \\ \\ = [r \ \frac{e^{-2r/a}}{(-2/a)}]^b__0}}- \int\limits^b_0 \frac{e^{-2r/a}}{(-2/a)}} \\ \\ \\ = \frac{ab \ e^{-2b/a}}{(-2)} + \frac{a}{2} \ \int\limits^b_a \ e^{-2r/a} \ dr$

$= \frac{ab \ e^{-2b/a}}{(-2)} + \frac{a}{2} \frac {[e^{-2r/a}]^b_0}{(-2/a)}$

$\int\limits^b_0 \ r \ e^{-2r/a} \ dr = \frac{ab \ e^{-2b/a}}{(-2)} - \frac{a^2}{4}[e^{-2b/a}-1]$

Replacing the value of $\int\limits^b_0 \ r \ e^{-2r/a} \ dr$ into equation (1); we have:

$Probability (P) = \frac{4}{a__0}^3}[(\frac{a_0b^2e^{-2b/a_0}}{(-2)})+ \frac{ab \ e^{-2b/a}}{(-2)} - \frac{a^2}{4}[e^{-2b/a}-1]]$

Substituting b = 1.1 a__0 ; we have:

$Probability (P) = \frac{4}{a^3}[(\frac{a^3(1.1)^2e^{-2(1.1)}}{(-2)})+ {a^2_3(1.1) \ e^{-2*1.1}} - \frac{a^3}{4}[e^{-2(1.1)}-1]]$

Probability (P) = 4 (-0.067036 - 0.06094 + 0.222299)

Probability (P) = 0.377292

Probability (P) ≅ 37.73 %

Answer from: coolkason

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:

P (r ) = r^2*R^2 ( r )

- 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.

$R ( r ) = 2*(\frac{1}{a_o} )^(^\frac{3}{2}^) * e^(^-^\frac{r}{a_o}^)$

Where, a_o = 52.9 pm , 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:

$P ( r ) = \int [ 2*(\frac{1}{a_o} )^(^\frac{3}{2}^) * e^(^-^\frac{r}{a_o}^) ] ^2*r^2 dr\\\\P ( r ) = \int [ 4*(\frac{1}{a_o} )^(^3^) * e^(^-^\frac{2r}{a_o}^) ] *r^2 . dr\\\\P ( r ) = 4*(\frac{1}{a_o} )^(^3^) \int [ e^(^-^\frac{2r}{a_o}^) . r^2 ] . dr$

- Perform integration by parts:

$P ( r ) = 4*(\frac{1}{a_o} )^(^3^) * ( [ \frac{e^(^-^\frac{2r}{a_o}^) }{\frac{-2}{a_o} } ]*r^2 - \int [ e^(^-^\frac{2r}{a_o}^) . 2r ] . dr)$

Answer from: Quest

Im pretty sure it would be 6 purple rectangles.

Answer from: Quest

i dont have an answer

step-by-step explanation:

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