Question

What is the average distance and the most probable distance of an electron from the nucleus in the 1s orbital of a hydrogen atom? ( \[{a_0}\] = the radius of the first Bohr orbit)
A.1.5 \[{a_0}\] and \[{a_0}\]
B.\[{a_0}\] and 5 \[{a_0}\]
C.1.5 \[{a_0}\]and 5 \[{a_0}\]
D. \[{a_0}\] and 0.5 \[{a_0}\]

Answer 1

Hint: The most probable distance is the distance where the electron is highest probable so it is the radius of that orbit and the average distance is the expectation value of the distance of an electron from the nucleus.

Complete step by step answer:
-The average distance is the expectation value of the distance of an electron from the nucleus i.e. if the radius of the first Bohr orbit is \[{a_0}\] , the average distance of an electron from the nucleus in the 1s orbital of a hydrogen atom can be calculated by the formula \[r = 0.529 \times \dfrac{{{n^2}}}{Z}\mathop A\limits^o \]. This is the expression for the radius of Bohr’s orbit in hydrogen and hydrogen like species. So, the average will be 3/2 \[{a_0}\] or 1.5 \[{a_0}\]. It is calculated by the following integral-
 \[\left\langle r \right\rangle = \int\limits_0^\infty {r\dfrac{{dP}}{{dr}}} dr = \dfrac{4}{{a_0^3}}\int\limits_0^\infty {{r^3}} {e^{ - 2r/{a_0}}}dr\]
Since all the terms containing r will be zero due to integration by parts, we get \[\left\langle r \right\rangle = \dfrac{{3{a_0}}}{2}\]. Thus proved.
-In order to find the maximum and minimum of a function, we usually take the derivative of the function with respect to the variable and set the derivative equal to zero. Similarly, to find the most probable distance of an electron we take the derivative of the probability density P(r) with respect to r and set it equal to zero.
 \[{P_{10}} = \dfrac{{4{r^2}}}{{a_0^3}}{e^{ - 2r/{a_0}}}\] this is the equation for the probability density for 1s orbital of a hydrogen atom. Taking its derivative with respect to r and setting it equal to zero results in,
 \[\dfrac{d}{{dr}}{P_{10}}(r) = 0 = \dfrac{d}{{dr}}\left( {\dfrac{{4{r^2}}}{{a_0^3}}{e^{ - 2r/{a_0}}}} \right)\]
 \[
  \dfrac{{2{r^2}}}{{{a_0}}} = 2r \\
  \therefore r = {a_0} \\
 \]
The most probable distance is the distance where the electron is highest probable so it is the radius of that orbit \[{a_0}\] which is equal to 0.529 \[\mathop A\limits^o \].

Hence, the correct option is (A).

Note:
The probability density accounts for the average being greater than the most probable value i.e. 1.5 times the most probable value. This is explained by Schrodinger equation as 1s orbital is spherically symmetrical and has no angular terms.