Question

a mixture of argon and mercury vapor are often used in advertising signs. the threshold value to emit visible light occurs at a wavelength of 470 nm.
calculate the energy change resulting from the emission of one photon at this wavelength.
$h = 6.626 \times 10^{-34}\\ j \cdot sec$
$c = 3.0 \times 10^8\\ m/s$
$? \times 10^{?}\\ j$

Explanation:

Step1: Convert wavelength to meters

$\lambda = 470\ \text{nm} = 470 \times 10^{-9}\ \text{m} = 4.7 \times 10^{-7}\ \text{m}$

Step2: Use photon energy formula

Recall $E = \frac{hc}{\lambda}$, substitute values:
$E = \frac{(6.626 \times 10^{-34}\ \text{J·sec})(3.0 \times 10^8\ \text{m/s})}{4.7 \times 10^{-7}\ \text{m}}$

Step3: Calculate numerator first

$(6.626 \times 10^{-34})(3.0 \times 10^8) = 1.9878 \times 10^{-25}$

Step4: Divide by wavelength

$E = \frac{1.9878 \times 10^{-25}}{4.7 \times 10^{-7}} \approx 4.23 \times 10^{-19}$

Answer:

$4.23 \times 10^{-19}\ \text{J}$