Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (Bohr's radius is represented by $$a_0$$)

To determine the incorrect statement, we evaluate each option using the quantum mechanical model of the hydrogen atom.

(A) The probability density of finding the electron is maximum at the nucleus.

This statement is correct. The probability density is given by $$|\psi|^2.$$

For an $$s$$-orbital, where the angular momentum quantum number is $$l=0,$$ the wavefunction does not vanish at the origin. Therefore, $$|\psi|^2$$ attains its maximum value at the nucleus, $$r=0.$$

It should not be confused with the radial probability distribution, $$4\pi r^2|\psi|^2,$$ which is zero at the nucleus and has its maximum value at the Bohr radius, $$a_0.$$

(B) The electron can be found at a distance $$2a_0$$ from the nucleus.

This statement is correct. The wavefunction decreases asymptotically with increasing distance and becomes zero only at infinity. Hence, there is a finite probability of finding the electron at any finite distance from the nucleus, including $$2a_0.$$

(C) The $$1s$$ orbital is spherically symmetrical.

This statement is correct. Since the $$1s$$ orbital has $$l=0,$$ its wavefunction has no angular dependence, giving rise to a perfectly spherical electron density distribution around the nucleus.

(D) The total energy of the electron is maximum when it is at a distance $$a_0$$ from the nucleus.

This statement is incorrect. In a stationary state such as the $$1s$$ orbital, the total energy of the electron remains constant and is independent of its position.

For the ground state of the hydrogen atom, $$E=-13.6\ \text{eV},$$ regardless of the instantaneous distance of the electron from the nucleus. Although the kinetic and potential energies vary with distance, their sum remains constant.

Hence, statement (D) is incorrect and is the correct answer.