Explanation Given: Wave function is given to be ψ 1 s ( ρ ) = 1 π a 0 3 e − ρ a 0 The probability of finding the electron in a region W of R 3 is equal to ∭ W p ( x , y , z ) d V Where, in spherical coordinates, p ( ρ ) = | ψ 1 s ( ρ ) | 2 Calculation: Since we have set up for probability as ∭ W p ( x , y , z ) d V , where p ( ρ ) = | ψ 1 s ( ρ ) | 2 for ψ 1 s ( ρ ) = 1 π a 0 3 e − ρ a 0 . Thus, we can write: ⇒ p ( ρ ) = | 1 π a 0 3 e − ρ a 0 | 2 ⇒ p ( ρ ) = e − 2 ρ a 0 π a 0 3 Thus, the triple integral for probability becomes: ⇒ ∭ W p ( x , y , z ) d V = 1 π a 0 3 ∭ W e − 2 ρ a 0 d V ⇒ ∭ W p ( x , y , z ) d V = 1 π a 0 3

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.5, Problem 58E

To determine

To prove:

That probability of finding the electron at a distance greater than Bohr’s radius is equal to $\frac{5}{e^{2}}$ .

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Chapter 16.5, Problem 57E

Chapter 16.5, Problem 59E

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Chapter 16 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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To prove: That probability of finding the electron at a distance greater than Bohr’s radius is equal to 5 e 2 .

Solution Summary: The author proves that probability of finding the electron at a distance greater than Bohr’s radius is equal to 5 e 2.

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To prove:

That probability of finding the electron at a distance greater than Bohr’s radius is equal to $\frac{5}{e^{2}}$ .

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Chapter 16 Solutions

To prove: That probability of finding the electron at a distance greater than Bohr’s radius is equal to 5 e 2 .

Solution Summary: The author proves that probability of finding the electron at a distance greater than Bohr’s radius is equal to 5 e 2.

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To prove: That probability of finding the electron at a distance greater than Bohr’s radius is equal to 5e2.

Want to see the full answer?

Chapter 16 Solutions

To prove:

That probability of finding the electron at a distance greater than Bohr’s radius is equal to $\frac{5}{e^{2}}$ .