a uv light with a wavelength of 125. nm is shone on the platinum surface. what is the incident energy from…

Explanation:

Step1: Convert wavelength to meters

$\lambda = 125\ \text{nm} = 125 \times 10^{-9}\ \text{m} = 1.25 \times 10^{-7}\ \text{m}$

Step2: Use photon energy formula

The formula for photon energy is $E = \frac{hc}{\lambda}$, where $h=6.626 \times 10^{-34}\ \text{J·s}$ (Planck's constant) and $c=3.00 \times 10^8\ \text{m/s}$ (speed of light). $$ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{1.25 \times 10^{-7}} $$

Step3: Calculate the numerator

$(6.626 \times 10^{-34})(3.00 \times 10^8) = 1.9878 \times 10^{-25}$

Step4: Divide by wavelength

$E = \frac{1.9878 \times 10^{-25}}{1.25 \times 10^{-7}} = 1.59024 \times 10^{-18}\ \text{J}$

Answer:

$E_I = 1.59 \times 10^{-18}\ \text{J}$