The wave function for the 2s orbital of hydrogen is given by Psi = (1/(4 sqrt(2))) (1/a0)^(3/2) (2 - r/a0) exp(-r/(2 a0)), where a0 = 0.529 Angstrom is the Bohr radius. Identify the correct statement(s).

Question

Accepted Answer

The probability density of finding the electron is independent of direction.. Radial nodes occur where the radial wavefunction equals zero (excluding infinity). Setting 2 - r/a0 = 0 gives r = 2a0: only one radial node, not three. The 2s orbital (l=0) has no angular part, so the probability density |Psi|² is spherically symmetric, i.e., independent of direction. At the nucleus (r=0): Psi = (1/(4 sqrt(2)))(1/a0)^(3/2)(2)(1) which is non-zero, so |Psi|² is non-zero at the nucleus. The radial node is at r=2a0. Correct statements: B, C, D.

The wave function for the 2s orbital of hydrogen is given by Psi = (1/(4 sqrt(2))) (1/a0)^(3/2) (2 - r/a0) exp(-r/(2 a0)), where a0 = 0.529 Angstrom is the Bohr radius. Identify the correct statement(s).

Solution

Related JEE Advanced Chemistry questions