Chapter 27, Problem 60P

To determine

0To Determine:

The value of n for a Bohr orbit and the energy of this orbit.

Answer to Problem 60P

Solution:

The value of n is $4.35\times {10}^3$ and energy associated with this state is $-7.2\times {10}^{-8}\text{eV}$.

Explanation of Solution

Given:

Radius of the hydrogen atom: $1.00\text{cm}$

Formula Used:

$\begin{array}{l}{r}_n=n^2{r}_1\\ E=-\frac{13.6}{n^2}\end{array}$

Calculation:

The value of the state can be calculated as

$\begin{array}{l}{r}_n=n^2{r}_1\\ 0.01=n^2(0.529\times {10}^{-10}\text{m})\\ n=13749\end{array}$

Now, by using the formula, $E=-\frac{13.6}{n^2}$

The value of the energy associated with this state can be calculated as

$E=-\frac{13.6}{n^2}=-\frac{13.6\text{eV}}{{(13749)}^2}=-7.2\times {10}^{-8}\text{eV}$