The first orbital of H is represented by :${\rm{\psi }} = \frac{1}{{\sqrt {\rm{\pi }} }}{\left( {\frac{1}{{{{\rm{a}}_0}}}} \right)^{3/2}}{{\rm{e}}^{ - {\rm{r}}/{{\rm{a}}_0}}},$ , where ${a_0}$ is Bohr's radius. The probability of finding the electron at a distance r, from the nucleus in the region dV is:

${{\rm{\psi }}^2}4{\rm{\pi }}{r^2}dr$

$\smallint {{\rm{\psi }}^2}4{\rm{\pi }}{r^2}d{\rm{v}}$

$\smallint {\rm{\psi }}d{\rm{v}}$

NEET UG CHEMISTRY MCQ: STRUCTURE OF ATOM

1 Hard

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Understanding +4 | -1

The first orbital of H is represented by :

\mathrm{\psi } = \frac{1}{{\sqrt \mathrm{\pi } }}{\left( {\frac{1}{{{\mathrm{a}_0}}}} \right)^{3/2}}{\mathrm{e}^{ - \mathrm{r}/{\mathrm{a}_0}}},

, where

{a_0}

is Bohr's radius. The probability of finding the electron at a distance r, from the nucleus in the region dV is:

View answer and explanation

Correct option: C

{\mathrm{\psi }^2}4\mathrm{\pi }{r^2}dr

Option C (

{\mathrm{\psi }^2}4\mathrm{\pi }{r^2}dr

) is correct because it represents the probability of finding the electron within a spherical shell of radius r and thickness dr.