Question

1. answer the following questions regarding a h atom in which an electron moves from the fourth energy level back to the second energy level. for both questions, express your answers to three significant figures.

a. calculate the energy change (in j) for the h atom. (7 pt)
δe = (-2.178×10⁻¹⁸ j)(1/(2)² - 1/(4)²)
δe = (-2.178×10⁻¹⁸ j)(1/4 - 1/16)
δe = (-2.178×10⁻¹⁸ j)(0.1875)
δe = -4.14×10⁻¹⁹ j 0.1875 3 s.f.

b. what is the wavelength (in nm) of each photon of light emitted given this energy change? (11 pt)

Explanation:

Step1: Recall energy - wavelength formula

The energy - wavelength relationship is $E = h
u=\frac{hc}{\lambda}$, where $E$ is energy, $h = 6.626\times10^{-34}\ J\cdot s$ (Planck's constant), $c= 2.998\times10^{8}\ m/s$ (speed of light), and $\lambda$ is wavelength. We know from part (a) that $E = 4.14\times 10^{-19}\ J$ (taking the magnitude since energy of the photon is positive).

Step2: Rearrange formula for wavelength

We can rearrange $E=\frac{hc}{\lambda}$ to $\lambda=\frac{hc}{E}$.

Step3: Substitute values

Substitute $h = 6.626\times10^{-34}\ J\cdot s$, $c = 2.998\times10^{8}\ m/s$, and $E = 4.14\times 10^{-19}\ J$ into the formula:
\[

$$\begin{align*} \lambda&=\frac{(6.626\times 10^{-34}\ J\cdot s)\times(2.998\times 10^{8}\ m/s)}{4.14\times 10^{-19}\ J}\\ &=\frac{6.626\times2.998\times10^{-34 + 8}}{4.14\times 10^{-19}}\ m\\ &=\frac{19.864748\times10^{-26}}{4.14\times 10^{-19}}\ m\\ &=\frac{1.9864748\times 10^{-25}}{4.14\times 10^{-19}}\ m\\ &= 4.80\times10^{-7}\ m \end{align*}$$

\]

Step4: Convert to nanometers

Since $1\ nm=10^{-9}\ m$, then $\lambda=(4.80\times 10^{-7}\ m)\times\frac{1\ nm}{10^{-9}\ m}=480\ nm$.

Answer:

$480\ nm$