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 $\Psi$ _1s(r) = 1/( $\pi a _ { 0 } ^ { 3 }$ ) ^1/2 exp(-r/a₀) where a₀ is the Bohr radius.The probability of finding the electron is

• A) \(\left( \pi / a _ { 0 } ^ { 3 } \right) \int \exp \left( - 2 r / a _ { 0 } \right) 4 \pi r ^ { 2 } d r\).
• B) \(\left( \pi / a _ { 0 } ^ { 3 } \right) ^ { 1 / 2 } \int \exp \left( - 2 r / a _ { 0 } \right) 4 \pi r ^ { 2 } d r\)
• C) \(\left( \pi / a _ { 0 } ^ { 3 } \right) ^ { 1 / 2 } \int \exp \left( - 2 r / a _ { 0 } \right) d r\)
• D) \(\left( \pi / a _ { 0 } ^ { 3 } \right) ^ { 1 / 2 } \int \exp \left( - 2 r / a _ { 0 } \right) d r\).
• E) \(\frac { d } { d r } \left[ \frac { 2 } { \pi a _ { 0 } ^ { 3 } } \exp \left( - 2 r / a _ { 0 } \right) \right] \left[ \frac { 1 } { \pi a _ { 0 } ^ { 3 } } \right] ^ { 1 / 2 } e ^ { - r / a ^ { 0 } 1 / 2 }\)

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