Question

metallic sodium has a photoelectric threshold value with a wavelength of 540. nanometers.
what is the energy change resulting from the emission of one photon at this wavelength?
$c = 3.0 \times 10^8 \text{ m/s}$ $h = 6.626 \times 10^{-34} \text{ j} \cdot \text{sec}$
$? \times 10^{?} \text{ j}$

Explanation:

Step1: Convert wavelength to meters

First, convert 540. nm to meters:
$540.\ \text{nm} = 540. \times 10^{-9}\ \text{m} = 5.40 \times 10^{-7}\ \text{m}$

Step2: Calculate photon energy

Use the formula $E = \frac{hc}{\lambda}$, where $h=6.626 \times 10^{-34}\ \text{J·sec}$, $c=3.0 \times 10^8\ \text{m/s}$, $\lambda=5.40 \times 10^{-7}\ \text{m}$:

$$ E = \frac{(6.626 \times 10^{-34}) \times (3.0 \times 10^8)}{5.40 \times 10^{-7}} $$

Step3: Compute the result

First calculate the numerator: $(6.626 \times 3.0) \times 10^{-34+8} = 19.878 \times 10^{-26}$
Then divide by the denominator: $\frac{19.878 \times 10^{-26}}{5.40 \times 10^{-7}} \approx 3.68 \times 10^{-19}\ \text{J}$

Answer:

$3.68 \times 10^{-19}\ \text{J}$