Solved

The Radial Portion of the De Broglie Wavefunction for an Electron

Question 28

Multiple Choice

The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is Ψ\Psi 1s(r) = 1/( πa03\pi a _ { 0 } ^ { 3 } ) 1/2 exp(-r/a0) where a0 is the Bohr radius.The probability of finding the electron is


A) (π/a03) exp(2r/a0) 4πr2dr\left( \pi / a _ { 0 } ^ { 3 } \right) \int \exp \left( - 2 r / a _ { 0 } \right) 4 \pi r ^ { 2 } d r .
B) (π/a03) 1/2exp(2r/a0) 4πr2dr\left( \pi / a _ { 0 } ^ { 3 } \right) ^ { 1 / 2 } \int \exp \left( - 2 r / a _ { 0 } \right) 4 \pi r ^ { 2 } d r
C) (π/a03) 1/2exp(2r/a0) dr\left( \pi / a _ { 0 } ^ { 3 } \right) ^ { 1 / 2 } \int \exp \left( - 2 r / a _ { 0 } \right) d r
D) (π/a03) 1/2exp(2r/a0) dr\left( \pi / a _ { 0 } ^ { 3 } \right) ^ { 1 / 2 } \int \exp \left( - 2 r / a _ { 0 } \right) d r .
E) ddr[2πa03exp(2r/a0) ][1πa03]1/2er/a01/2\frac { d } { d r } \left[ \frac { 2 } { \pi a _ { 0 } ^ { 3 } } \exp \left( - 2 r / a _ { 0 } \right) \right] \left[ \frac { 1 } { \pi a _ { 0 } ^ { 3 } } \right] ^ { 1 / 2 } e ^ { - r / a ^ { 0 } 1 / 2 }

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions