# Question #27e1c

Jul 19, 2017

$0$

#### Explanation:

Your goal here is to figure out if $\text{9.9 eV}$ of energy are enough to excite a hydrogen atom in its ground state and if so, the highest energy level that the electron can reach.

The easiest way you have of figuring out if $\text{9.9 eV}$ is enough energy to bump the electron from the first energy level is to subtract the energy of the ground level, for which $n = 1$, and the energy of the first excited state, for which $n = 2$.

You will end up with--you can ignore the minus signs for this purpose

$\text{13.6 eV " - " 3.40 eV" = "10.2 eV}$

This tells you that in order for an electron to move from $n = 1$ to $n = 2$ in a hydrogen atom, it must absorb $\text{10.2 eV}$ of energy.

Since you only have $\text{9.9 eV}$ available, you can say that the electron won't absorb the energy needed for this transition to take place, which implies that the number of spectral lines emitted will be equal to $0$.

Alternatively, you can use the fact that the quantized energy levels that the electron can occupy in a hydrogen atom are described by the following equation

${E}_{n} = - \frac{\text{13.6 eV}}{n} ^ 2$

Here $n$ represents the principal quantum number that describes a given energy level.

In your case, you will have

${E}_{n} = - \text{13.6 eV" + "9.9 eV" = -"3.7 eV}$

This means that $n$ will be

${n}^{2} = \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{-}}} 13.6 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{eV"))))/(color(red)(cancel(color(black)(-)))3.7color(red)(cancel(color(black)("eV}}}}\right) \implies n = \sqrt{\frac{13.6}{3.7}} \approx 1.9$

Since you have $1 < n < 2$, you can conclude that the electron did not absorb enough energy to get to the second energy level $\to$ no spectral lines will be emitted.