Question

A photon has energy of \(3.1\times10^{-19}\ \text{J}\). Its wavelength (in \(\AA\)) is _ _ _. (round off to one decimal place)

Hint:

For photons: \[ E=\frac{hc}{\lambda} \] Higher photon energy corresponds to shorter wavelength.

Answer 1

Correct Answer: 6387.1

Step 1: Write the photon relation.
Energy and wavelength link by \[ E = \frac{hc}{\lambda} \] so $\lambda = \dfrac{hc}{E}$.

Step 2: Gather the constants.
Take $h = 6.6\times 10^{-34}$ J s, $c = 3\times 10^{8}$ m/s, and the given $E = 3.1\times 10^{-19}$ J.

Step 3: Plug in.
\[ \lambda = \frac{(6.6\times10^{-34})(3\times10^{8})}{3.1\times10^{-19}} = \frac{19.8\times10^{-26}}{3.1\times10^{-19}} \]

Step 4: Compute and convert.
This gives about $6.387\times10^{-7}$ m. Since $1\ \text{\AA} = 10^{-10}$ m, that is about $6387.1\ \text{\AA}$.

Step 5: Answer.
\[ \boxed{6387.1\ \text{\AA}} \]