Explanation Given Information: 2 s and 2 p state Formula Used: The expression for the uncertainty in r is as follows: Δ r = r 2 ¯ − r ¯ 2 Calculation: For 2 s state: n = 2 and l = 0 the radial solution for the 2 s state is as follows. R 2 s = 2 ( 2 a 0 ) 3 2 ( 1 − r 2 a 0 ) e − r 2 a 0 For 2 p state: n = 2 and l = 1 the radial solution for the 2 s state is as follows. R 2 p = 2 ( 2 a 0 ) 3 2 ( r 3 a 0 ) e − r 2 a 0 Here, a 0 is the Bohr radius and its value is a 0 = 0.0529 n m For 2 s state The expression for the expectation value of r is as follows. r ¯ = ∫ 0 ∞ r P ( r ) d r Here P ( r ) is the probability per unit radial distance and the value of P ( r ) is P ( r ) = r 2 R 2 ( r ) Substitute 2 ( 2 a 0 ) 3 2 ( 1 − r 2 a 0 ) e − r 2 a 0 for R ( r ) in equation P ( r ) = r 2 R 2 ( r ) P ( r ) = r 2 ( 2 ( 2 a 0 ) 3 2 ( 1 − r 2 a 0 ) e − r 2 a 0 ) 2 = r 2 2 a 0 3 ( 1 − r a 0 + r 2 4 a 0 2 ) e − r a 0 = 1 2 a 0 3 ( r 2 − r 3 a 0 + r 4 4 a 0 2 ) e − r a 0 Substitute 1 2 a 0 3 ( r 2 − r 3 a 0 + r 4 4 a 0 2 ) e − r a 0 for P ( r ) in equation r ¯ = ∫ 0 ∞ r P ( r ) d r and solve for r ¯ . r ¯ = ∫ 0 ∞ ( r ) 1 2 a 0 3 ( r 2 − r 3 a 0 + r 4 4 a 0 2 ) e − r a 0 d r = 1 2 a 0 3 ∫ 0 ∞ ( r 3 − r 4 a 0 + r 5 4 a 0 2 ) e − r a 0 d r = 1 2 a 0 3 ( 3 ! ( 1 / a 0 ) 4 − 1 a 0 ( 4 ! ( 1 / a 0 ) 5 ) + 1 4 a 0 ( 5 ! ( 1 / a 0 ) 6 ) ) = 6 a 0 The expression for the expectation value of r is as follows. r 2 ¯ = ∫ 0 ∞ r 2 P ( r ) d r Substitute 1 2 a 0 3 ( r 2 − r 3 a 0 + r 4 4 a 0 2 ) e − r a 0 for P ( r ) in equation r 2 ¯ = ∫ 0 ∞ r 2 P ( r ) d r and solve for r 2 ¯ . r 2 ¯ = ∫ 0 ∞ ( r 2 ) 1 2 a 0 3 ( r 2 − r 3 a 0 + r 4 4 a 0 2 ) e − r a 0 d r = 1 2 a 0 3 ∫ 0 ∞ ( r 4 − r 5 a 0 + r 6 4 a 0 2 ) e − r a 0 d r = 1 2 a 0 3 ( 4 ! ( 1 / a 0 ) 5 − 1 a 0 ( 5 ! ( 1 / a 0 ) 6 ) + 1 4 a 0 ( 6 ! ( 1 / a 0 ) 7 ) ) = 42 a 0 2 Calculate the uncertainty in r for 2 s state as follows: The expression for the uncertainty in r is as follows: Δ r = r 2 ¯ − r ¯ 2 Substitute 42 a 0 2 for r 2 ¯ and 6 a 0 for r ¯ in equation Δ r = r 2 ¯ − r ¯ 2 and solve for Δ r

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Author: Randy Harris

Publisher: Addison Wesley

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Chapter 7, Problem 62E

To determine

To Calculate:

The uncertainties in r for the 2s and 2p states.

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Chapter 7, Problem 61E

Chapter 7, Problem 63E

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Consider the situation in the figure below; a neutral conducting ball hangs from the ceiling by an insulating string, and a charged insulating rod is going to be placed nearby. A. First, if the rod was not there, what statement best describes the charge distribution of the ball? 1) Since it is a conductor, all the charges are on the outside of the ball. 2) The ball is neutral, so it has no positive or negative charges anywhere. 3) The positive and negative charges are separated from each other, but we don't know what direction the ball is polarized. 4) The positive and negative charges are evenly distributed everywhere in the ball. B. Now, when the rod is moved close to the ball, what happens to the charges on the ball? 1) There is a separation of charges in the ball; the side closer to the rod becomes positively charged, and the opposite side becomes negatively charged. 2) Negative charge is drawn from the ground (via the string), so the ball acquires a net negative charge. 3)…

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Modern Physics

Ch. 7 - Prob. 1CQ Ch. 7 - Prob. 2CQ Ch. 7 - Prob. 3CQ Ch. 7 - Prob. 4CQ Ch. 7 - Prob. 5CQ Ch. 7 - Prob. 6CQ Ch. 7 - Prob. 7CQ Ch. 7 - Prob. 8CQ Ch. 7 - Prob. 9CQ Ch. 7 - What are the dimensions of the spherical harmonics...

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