Chapter 27, Problem 48P

(a)

To determine

The difference in radius between the $n=1$ and $n=2$ state for hydrogen.

Answer to Problem 48P

The difference in radius between the $n=1$ and $n=2$ state for hydrogen is $\underset{\_}{1.59\times {10}^{-10}\text{m}}$.

Explanation of Solution

Write the expression for the radius of $n^{\text{th}}$ state of hydrogen atom.

  $r_n=n^2{a}_0$                                                                                                                      (I)

Here, $r_n$ is the radius of the $n^{\text{th}}$ state orbit, and $a_0$ is the Bohr radius ($a_0=52.9\times {10}^{-12}\text{m}$).

Use equation (I) to write the radius of the $n=1$ and $n=2$ state.

  $r_1=1^2{a}_0$                                                                                                              (II)

  $r_2=2^2{a}_0$                                                                                                           (III)

Write the expression for the difference in radius between the $n=1$ and $n=2$ state for hydrogen using equation (II) and (III).

  $\begin{array}{c}{r}_2-r_1=2^2{a}_0-1^2{a}_0\\ =3a_0\end{array}$                                                                                           (IV)

Conclusion:

Substitute $52.9\times {10}^{-12}\text{m}$ for $a_0$ in equation (IV) to find the difference in radius between the $n=1$ and $n=2$ state for hydrogen.

  $\begin{array}{c}{r}_2-r_1=3\left(52.9\times {10}^{-12}\text{m}\right)\\ =1.59\times {10}^{-10}\text{m}\end{array}$

Therefore, the difference in radius between the $n=1$ and $n=2$ state for hydrogen is $\underset{\_}{1.59\times {10}^{-10}\text{m}}$.

(b)

To determine

The difference in radius between the $n=100$ and $n=101$ state for hydrogen, and the variation in orbital separation for large and small values of $n$.

Answer to Problem 48P

The difference in radius between the $n=100$ and $n=101$ state for hydrogen is $\underset{\_}{1.06\times {10}^{-8}\text{m}}$. The orbital separations are much larger for larger $n$ values.

Explanation of Solution

Use equation (I) to write the radius of the $n=100$ and $n=101$ state.

  $r_{100}={100}^2{a}_0$                                                                                                       (V)

  $r_{101}={101}^2{a}_0$                                                                                                       (VI)

Write the expression for the difference in radius between the $n=100$ and $n=101$ state for hydrogen using equation (V) and (VI).

  $\begin{array}{c}{r}_{101}-r_{100}={101}^2{a}_0-{100}^2{a}_0\\ =\left({101}^2-{100}^2\right)a_0\end{array}$                                                                               (VII)

Conclusion:

Substitute $52.9\times {10}^{-12}\text{m}$ for $a_0$ in equation (VII) to find the difference in radius between the $n=100$ and $n=101$ state for hydrogen.

  $\begin{array}{c}{r}_{101}-r_{100}=\left({101}^2-{100}^2\right)\left(52.9\times {10}^{-12}\text{m}\right)\\ =1.06\times {10}^{-8}\text{m}\end{array}$

This indicates that the orbital separation of hydrogen increases as the value of $n$ increases. For small values of $n$, the orbital separation is very small. For larger values of $n$, the orbital separation is also large.

Therefore, the difference in radius between the $n=100$ and $n=101$ state for hydrogen is $\underset{\_}{1.06\times {10}^{-8}\text{m}}$. The orbital separations are much larger for larger $n$ values.