Answer
The wavelength of this radiation is $327.8nm$.
Work Step by Step
*Strategy:
This exercise is actually a reverse of the exercise in part b). Therefore, we would employ the same formulas, only in reversed order.
- First, we would use the Planck's theory formula $E=h\times\nu$ to find the frequency of radiation.
- Then, we use the formula $\lambda\times\nu = c$ to find the wavelength of this radiation.
$\lambda$ : wavelength of radiation
$\nu$ : frequency of radiation
$c$ : speed of light in a vacuum ($c\approx2.998\times10^8m/s$)
$E$ : the energy of a single quantum (or in this case, photon)
$h$ : Planck constant ($h\approx6.626\times10^{-34}J.s$)
Step 1: Find the known variables
We know each photon of this radiation has an energy of $6.06\times10^{-19}J$; so $E=6.06\times10^{-19}J$.
Step 2: Find the frequency of this radiation
$\nu=\frac{E}{h}=\frac{6.06\times10^{-19}}{6.626\times10^{-34}}\approx9.146\times10^{14}s^{-1}$
Step 3: Find the wavelength of radiation
$\lambda=\frac{c}{\nu}=\frac{2.998\times10^8}{9.146\times10^{14}}\approx3.278\times10^{-7}m\approx327.8nm$
