Question

determine the energy, in j, of a photon with a wavelength of 373 nm. (h = 6.626×10⁻³⁴ j·s and c = 3.00×10⁸ m/s)

Explanation:

Step1: Convert wavelength to meters

Convert 373 nm to m. Since 1 nm = 10^{-9} m, then $\lambda=373\times10^{-9}$ m.

Step2: Use the energy - wavelength 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 $\lambda=373\times10^{-9}\ m$ into the formula.
$E=\frac{6.626\times10^{-34}\times3.00\times10^{8}}{373\times10^{-9}}$

Step3: Calculate the energy

$E=\frac{19.878\times10^{-26}}{373\times10^{-9}}=\frac{19.878}{373}\times10^{-26 + 9}\ J\approx0.0533\times10^{-17}\ J = 5.33\times10^{-19}\ J$

Answer:

$5.33\times10^{-19}$