Q11P Calculate the lifetime (in secon... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 9: Q11P (page 359)

Calculate the lifetime (in seconds) for each of the four n = 2 states of hydrogen. Hint: You鈥檒l need to evaluate matrix elements of the form $< 蠄_{100} |x| 蠄_{200} >, < 蠄_{100} |y| 蠄_{211} >$ , and so on. Remember that role="math" localid="1658303993600" $x = r 蝉颈苍胃肠辞蝉蠒, y = r 蝉颈苍胃蝉颈苍蠒 a n d z = r 肠辞蝉胃$ . Most of these integrals are zero, so inspect them closely before you start calculating. Answer: $1.60 脳 10^{- 9}$ seconds for all except role="math" localid="1658304185040" $蠄_{200}$ , which is infinite.

Short Answer

Expert verified

The life time for each of the four n=2 states of hydrogen is $蟿 = \frac{1}{A} = 1.60 脳 10^{- 9} s$

Step by step solution

Step 1: states of hydrogen

At standard conditions of hydrogen is a gas of diatomic molecules having the formula H2. It is colourless, tasteless, nontoxic, and highly combustible. Hydrogen is the most abundant chemical substance in the universe, constituting roughly 75% of all normal matter

Step 2: Calculating the life time for each of the four n=2 states of hydrogen

In Problem 9.1 we calculated the matrix elements of z; all of them are zero except $<100 |z| 210> = \frac{2^{8}}{3^{5} \sqrt{2}} a$ .

As for x and y, we noted that

$|100 >, 200| and 210 > are even (in x, y), whereas 21 + 1 >$ is odd. So the only non-zero matrix elements are $<100 |x| 21 + 1> and <100 |y| 21 + 1>$ Using the wave functions in Problem 9.1:

<100 |x| 21 + 1> = \frac{1}{\sqrt{{蟺补}^{3}}} (\frac{+ 1}{8 \sqrt{蟺补}}) \frac{1}{a} 鈭� e^{- r / a} r e^{- r / 2 a} \sin 胃 e^{+ i 蠒} (r \sin 胃肠辞蝉 蠒) r^{2} 蝉颈苍胃 d r d 胃 d 蠒 = + \frac{1}{8 {蟺补}^{4}} {鈭�}_{0}^{鈭�} r^{4} e^{- 3 r / 2 a} d r {鈭�}_{0}^{蟺} \sin^{3} 胃 d 胃 {鈭�}_{0}^{2 蟺} (肠辞蝉蠒 + i \sin 蠒) 肠辞蝉蠒 d 蠒 = \frac{+ 1}{8 {蟺补}^{4}} [4! {(\frac{2 a}{3})}^{5}] (\frac{4}{3}) (蟺) = + \frac{2^{7}}{3^{5}} a

$<100 |y| 21 + 1> = \frac{+ 1}{8 {蟺补}^{4}} [4! {(\frac{2 a}{3})}^{5}] (\frac{4}{3}) {鈭�}_{0}^{2 蟺} (\cos 蠒 + i \sin 蠒) \sin 蠒 d 蠒 = \frac{+ 1}{8 {蟺补}^{4}} [4! {(\frac{2 a}{3})}^{5}] (\frac{4}{3}) (+ i 蟺) = - i \frac{2^{7}}{3^{5}} a <100 |r| 200> = 0; <100 |r| 210> \frac{2^{7} \sqrt{2}}{3^{5}} a \hat{k}; <100 |r| 21 + 1> = \frac{2^{7}}{3^{5}} a (+ \hat{i} - \hat{i j}), and |21 + 1 > 鈫�| 100 >)$

Meanwhile, $蝇 = \frac{E_{2} - E_{1}}{魔} = \frac{1}{魔} (\frac{E_{1}}{4} - E_{1}) = - \frac{3 E_{1}}{4 魔}$ so for the three l = 1 states:

$A = - \frac{3^{3} E_{1}^{3}}{2^{6} 魔^{3}} \frac{{(e a)}^{2} 2^{15}}{3^{10}} \frac{1}{3 蟺 {\overset{藱}{o}}_{0} {魔肠}^{3}} = - \frac{2^{9}}{3^{8} 蟺} \frac{E_{1}^{3} e^{2} a^{2}}{{\overset{藱}{o}}_{0} 魔^{4} c^{3}} = \frac{2^{10}}{3^{8}} {(\frac{E_{1}}{m c^{2}})}^{2} \frac{c}{a} = \frac{2^{10}}{3^{8}} (\frac{13.6}{0.511 脳 10^{6}}) \frac{(3.00 脳 10^{8})}{(0.529 脳 10^{- 10} m)} = 6.27 脳 10^{8} / s; 蟿 = \frac{1}{A} = 1.60 脳 10^{- 9} s$

Thus life time for each of the four n=2 states of hydrogen is $蟿 = \frac{1}{A} = 1.60 脳 10^{- 9} s$

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Short Answer

Step by step solution

Step 1: states of hydrogen

Step 2: Calculating the life time for each of the four n=2 states of hydrogen

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