The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is Ψ₁_s(r) = 1/(
) ^1/2 exp(−r/a₀) where a₀ is the Bohr radius.The probability of finding the electron is

Question

The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is Ψ₁_s(r) = 1/(

) ^1/2 exp(−r/a₀) where a₀ is the Bohr radius.The probability of finding the electron is

Quizplus · Accepted Answer

The probability of finding the electron is given by the square of the absolute value of the wavefunction. Therefore,

P(r) = |Ψ(r)|^2

Substituting the given wavefunction,

P(r) = (1/(4πa0^3)) (1/2π) (1/a0)^(3/2) e^(-2r/a0)

Integrating this expression over all space (0 to infinity), we get

∫P(r) dV = (1/(4πa0^3)) (1/2π) (4π/3)(a0)^3

= 1/3

Therefore, the probability of finding the electron in the ground state of hydrogen atom is 1/3 or approximately 0.333.

Converting this to scientific notation,

P(r) = 11ea7ff9_2c6d_0e0a_a768_e3c7be3071b6_TB2295_11

Therefore, the best choice is A.

The Radial Portion of the De Broglie Wavefunction for an Electron