## Related questions with answers

A hydrogen atom is in its first excited state (n = 2). Using the Bohr theory of the atom, calculate the kinetic energy.

Solutions

VerifiedIn this problem, a hydrogen atom is excited to the first excited state, corresponding to $n = 2$. We use the Bohr model in this problem. For this part, we calculate the kinetic energy of the electron.

Given values:

$n=2$

In part $\textbf{d)}$ we have to find the kinetic energy.

For the $n=2$ state, kinetic energy $K.E_{n=2}$ can be calculated as:

$\begin{align*} K.E_{n=2}=\frac{m\upsilon^{2}}{2} \end{align*}$

$\begin{align*} \therefore{K.E_{n=2}}=\frac{{p_{n=2}}^{2}}{2m} \end{align*}$

Substituting initial value for the linear momentum, from the solution $\textit{11(b)}$, we get:

$\begin{equation} K.E_{n=2}=\frac{\left[9.95\times10^{-25}\frac{\textrm{kg}\cdot{\textrm{m}}}{\textrm{s}}\right]^{2}}{2\left[9.11\times10^{-31}\textrm{kg}\right]}=\frac{ 5.44\times10^{-19}\textrm{J}\left[1\textrm{eV}\right]}{1.60\times10^{-19}\textrm{J}} \end{equation}$

Solution is:

$\begin{align*} \boxed{K.E_{n=2}=3.4\textrm{eV}} \end{align*}$

$KE_2 = \dfrac{p_2^2}{2 \ m}$

$= \dfrac{(9.95e-25)^2}{(2) \ (9.11e-31)}$

$= 5.44e-19 \ J$

$= 3.40 \ eV$

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Mathematical Methods in the Physical Sciences

3rd Edition•ISBN: 9780471198260 (1 more)Mary L. Boas#### Fundamentals of Physics

10th Edition•ISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick## More related questions

1/4

1/7