Question

Calculate the lifetime (in seconds) for each of the four n = 2 states of hydrogen. Hint: You’ll need to evaluate matrix elements of the form $<{\psi }_{100}\left|x\right|{\psi }_{200}>,<{\psi }_{100}\left|y\right|{\psi }_{211}>$, and so on. Remember that role="math" localid="1658303993600" $\mathit{x}\mathbf{=}\mathit{r}\mathbf{}\mathbf{sin\theta cos}\mathbf{}\mathit{\varphi }\mathbf{,}\mathit{y}\mathbf{=}\mathit{r}\mathbf{}\mathbf{sin\theta sin}\mathbf{}\mathit{\varphi }\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{z}\mathbf{=}\mathit{r}\mathbf{}\mathbf{cos\theta }$. Most of these integrals are zero, so inspect them closely before you start calculating. Answer: $\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}$seconds for all except role="math" localid="1658304185040" ${\mathbf{\psi }}_{\mathbf{200}}$, which is infinite.

Short answer

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

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 .

As for x and y, we noted that

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

Meanwhile, $\omega =\frac{E_2-E_1}{ħ}=\frac{1}{ħ}\left(\frac{E_1}{4}-E_1\right)=-\frac{3E_1}{4ħ}$ so for the three l = 1 states:

$$\begin{gathered} A=-\frac{3^3{E}_1^3}{2^6{ħ}^3}\frac{{\left(ea\right)}^2{2}^{15}}{3^{10}}\frac{1}{3\mathrm{\pi }{\dot{\mathrm{o}}}_0{\mathrm{ħc}}^3} \\ =-\frac{2^9}{3^8\mathrm{\pi }}\frac{E_1^3{e}^2{a}^2}{{\dot{o}}_0{ħ}^4{c}^3} \\ =\frac{2^{10}}{3^8}{\left(\frac{E_1}{m{c}^2}\right)}^2\frac{c}{a} \\ =\frac{2^{10}}{3^8}\left(\frac{13.6}{0.511\times {10}^6}\right)\frac{\left(3.00\times {10}^8\right)}{\left(0.529\times {10}^{-10}m\right)} \\ =6.27\times {10}^8/s; \\ \tau =\frac{1}{A} \\ =1.60\times {10}^{-9}s \end{gathered}$$

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