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The Radial Portion of the De Broglie Wavefunction for an Electron

Question 12

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/() 1/2 exp(-r/a0) where a0 is the Bohr radius.The probability of finding the electron is: $The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is \Psi 1s(r) = 1/() 1/2 exp(-r/a0) where a0 is the Bohr radius.The probability of finding the electron is: A) B) C) D) E)$

A) $The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is \Psi 1s(r) = 1/() 1/2 exp(-r/a0) where a0 is the Bohr radius.The probability of finding the electron is: A) B) C) D) E)$
B) $The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is \Psi 1s(r) = 1/() 1/2 exp(-r/a0) where a0 is the Bohr radius.The probability of finding the electron is: A) B) C) D) E)$
C) $The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is \Psi 1s(r) = 1/() 1/2 exp(-r/a0) where a0 is the Bohr radius.The probability of finding the electron is: A) B) C) D) E)$
D) $The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is \Psi 1s(r) = 1/() 1/2 exp(-r/a0) where a0 is the Bohr radius.The probability of finding the electron is: A) B) C) D) E)$
E) $The radial portion of the de Broglie wavefunction for an electron in the ground state of the hydrogen atom is \Psi 1s(r) = 1/() 1/2 exp(-r/a0) where a0 is the Bohr radius.The probability of finding the electron is: A) B) C) D) E)$