Question

determine the energy of a photon, in j, with a wavelength of 424 nm. (h = 6.626×10^(-34) j·s and c = 3.00×10^8 m/s)
answer: j

Explanation:

Step1: Convert wavelength to meters

The wavelength $\lambda = 424\ nm=424\times10^{- 9}\ m$.

Step2: Use the photon - energy formula

The energy of a photon is given by $E = h
u$, and since $
u=\frac{c}{\lambda}$, then $E=\frac{hc}{\lambda}$. Substitute $h = 6.626\times10^{-34}\ J\cdot s$, $c = 3.00\times10^{8}\ m/s$ and $\lambda=424\times10^{-9}\ m$ into the formula.
$E=\frac{6.626\times10^{-34}\times3.00\times10^{8}}{424\times10^{-9}}$
$E=\frac{19.878\times10^{-26}}{424\times10^{-9}}$
$E=\frac{19.878}{424}\times10^{-26 + 9}$
$E\approx0.0469\times10^{-17}$
$E = 4.69\times10^{-19}\ J$

Answer:

$4.69\times10^{-19}\ J$