Answer
- The energy of photon of wavelength $3.3\mu m$ is $6.02\times10^{-20}J$.
- The energy of photon of wavelength $0.154nm$ is $1.29\times10^{-15}J$.
- The energy of photon of wavelength $0.154nm$ is higher than the energy of photon of wavelength $3.3\mu m$.
Work Step by Step
*Strategy:
We would calculate the energy of a photon of each wavelength according to the following formula $$E=h\times\nu=\frac{h\times c}{\lambda}$$
then compare the energy in each case with each other.
$E$ : the energy of a photon
$h$ : Planck constant ($h\approx6.626\times10^{-34}J.s$)
$\nu$ : frequency of photon
$\lambda$ : wavelength of photon
$c$ : speed of light in a vacuum ($c\approx2.998\times10^8m/s$)
- Step 1: Find the known variables
There is one photon (let's call it $p1$) of wavelength $3.3\mu m$, which means $\lambda_{p1} = 3.3\mu m= 3.3\times10^{-6}m$.
The other photon ($p2$) has wavelength $0.154nm$; so $\lambda_{p2}= 0.154nm=1.54\times10^{-10}m$.
- Step 2: Calculate the energy of photon $p1$ and photon $p2$
$E_{p1}=\frac{h\times c}{\lambda}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{3.3\times10^{-6}}\approx6.02\times10^{-20}J$
$E_{p2}=\frac{h\times c}{\lambda}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{1.54\times10^{-10}}\approx1.29\times10^{-15}J$
We see that the energy of photon $p2$ (wavelength $0.154nm$) is higher than the energy of photon $p1$ (wavelength $3.3\mu m$).
