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# 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 ${a_0}$

Last updated date: 13th Jun 2024
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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.

-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.
${P_{10}} = \dfrac{{4{r^2}}}{{a_0^3}}{e^{ - 2r/{a_0}}}$ 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 $\mathop A\limits^o$.