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

first option

a function is a relation in which an input is paired with exactly one output.

so, let's find the right answer.

in our first option, our inputs are january, february, and march. the outputs are 30, 28, and 25. each input has it's own output. this is correct!

our second option we know is incorrect, because ms. valverade has two outputs, dogs and cats. this is not a function.

our third option is incorrect as well. the yellow square has two ouputs, the square and the triangle.

lastly, in our fourth option, we see the number 7 two times with a different output. therefore, this is not a function.

remember: each input must be paired with only one output!

Me? ! with this question which relationship is a function?

Answer from: Quest

sorry, i can't understand your question. will later if you describe more clearly!

: )

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