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

Explanation:

Step1: Recall the energy - wavelength relationship

The energy of a photon is given by $E = h
u=\frac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\ J\cdot s$ (Planck's constant), $c = 3.0\times10^{8}\ m/s$ (speed of light), and $\lambda$ is the wavelength. We can rearrange the formula to solve for $\lambda=\frac{hc}{E}$.

Step2: Convert energy to SI units if necessary

The energy values are already in joules.

Step3: Calculate the wavelength for $\Delta E=- 4.09\times10^{-19}\ J$

Substitute $h = 6.626\times10^{-34}\ J\cdot s$, $c = 3.0\times10^{8}\ m/s$ and $E = 4.09\times10^{-19}\ J$ into $\lambda=\frac{hc}{E}$.
$\lambda=\frac{6.626\times10^{-34}\times3.0\times10^{8}}{4.09\times10^{-19}}=\frac{19.878\times10^{-26}}{4.09\times10^{-19}}\ m$.
$\lambda = 4.86\times10^{-7}\ m$.
Since $1\ nm=10^{-9}\ m$, $\lambda = 486\ nm$.

Answer:

486 nm