# Solve the probability of finding a 1 s electron at 1

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Solve: The probability of finding a 1 s electron at 1 B 2 r a < is ( ) ( ) ( ) B B B B 2 2 2 2 1 B 1 2 3 0 0 0 B 1 1 1 2 2 2 4 Prob 4 a a a r a r s r a P r dr r R r dr r e dr a π < = = = where we used the expression for ( ) 1 s R r found in Equation 42.7. Change the variable to u = 2 r/a B . This means 2 r dr = ( ) 3 2 8 . B a u du Also, the upper limit of the integral changes to u = 1 when r 1 B 2 a = . The probability is ( ) ( ) 3 1 1 2 2 B 1 B 2 3 0 0 B 1 2 1 1 0 4 1 Prob 8 2 1 1 5 2 2 5 2 1 0.0803 8.03% 2 2 2 u u u a r a u e du u e du a u u e e e < = = = + + = + = = =

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42.54. Solve: The radial probability density for the 2 s state is ( ) ( ) B B 2 3 4 2 2 2 2 2 2 2 2 B B B 4 4 1 2 4 r a r a r s s r r r P r r R r A r e C r e a a a π π = = = + where 2 2 4 s C A π = is a constant. This is a maximum where dP r /dr = 0. The derivative is B B B 2 3 3 4 2 2 2 B B B B B 2 3 2 3 B B B 3 1 2 4 4 2 2 4 r a r a r r a dP r r r r C r e r e dr a a a a a r r r C re a a a = + + = + The derivative is zero, giving the maxima and minima of P r ( r ), when 2 3 B B B 1 2 4 2 0 4 r r r a a a + = This is a cubic equation, which will have three roots. This is expected because we know from the graph in Figure 42.8 that P r ( r ) has not only two maxima but also, in between, a minimum. Although the cubic equation can be solved numerically, here we just need to verify that r = 5.236 a B is a solution. Substituting into the equation, 2 1 4 2 4(5.236) 2(5.236) + = 3 (5.236) 0. = Thus B 5.236 r a = is the most probable distance of the electron in the 2 s state.
42.55. Solve: The radial probability density for the 2 s state is ( ) ( ) B B 2 3 4 2 2 2 2 2 2 2 2 B B B 4 4 1 2 4 r a r a r s s r r r P r r R r A r e C r e a a a π π = = = + where 2 2 4 s C A π = is a constant. This is a maximum where dP r /dr = 0. The derivative is B B B 2 3 3 4 2 2 2 B B B B B 2 3 2 3 B B B 3 1 2 4 4 2 2 4 r a r a r r a dP r r r r C r e r e dr a a a a a r r r C re a a a = + + = + The derivative is zero, giving the maxima and minima of P r ( r ), when 2 3 B B B 1 2 4 2 0 4 r r r a a a + = This is a cubic equation, which will have three roots. This is expected because we know from the graph of Figure 42.8 that P r ( r ) has not only two maxima but also, in between, a minimum. The cubic equation can be solved numerically to give r = 0.764 a B , 2.00 a B , and 5.236 a B . By referring to the graph, we see that the outer two roots correspond to the maxima and the middle root is the minimum. The separation between the peaks is ( ) B B B 5.236 0.764 4.472 r a a a Δ = =

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42.56.
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• Winter '10
• E.Salik

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