Calculate the energy of a photon of electromagnetic radiation at each of the wavelengths indicated in Problem 39. a. 632.8 nm (wavelength of red light from helium–neon laser) b. 503 nm (wavelength of maximum solar radiation) c. 0.052 nm (wavelength contained in medical X-rays)

Video transcript

Hey everyone in this example, we're told that the wavelength of orange light is 5 86.9 nanometers. And we need to calculate the energy of a photon of this light. So we should recall that our formula for energy is going to be equal to Planck's constant, multiplied by our speed of light divided by our given wavelength. However, were given our units of wavelength in nanometers. And we want to go ahead and convert this to meters. So we should go ahead and find our energy calculation by again in a new meter, recalling that plank's constant is a value of 6.626 times 10 to the negative 34th power in units of jewels, times seconds. And then we're going to continue on and plug in our speed of light, which we recall is 3.0 times 10 to the eighth power in units of meters per second. In our denominator we're going to plug in that given wavelength. So we're given our wavelength represented by lambda as 86.9 nanometers. But we want to go ahead and cancel our units of nanometers. So we're going to multiply by the conversion factor. We're in our denominator. We should have nanometers And in our numerator we're going to have meters, we recall that our prefix nano tells us we have for one nanometer 10 to the negative ninth power meters. So now that we have nanometers aligned in the numerator and denominator, we can cancel them out leaving us with meters. And because we also have meters in our numerator, we can cancel out meters as well as we can cancel out the inverse seconds with the seconds from plank's constant. This leaves us with jewels as our final unit for energy here and what we're going to get for our value is a value equal to 3.38 times 10 to the negative 19th power jewels. And this is per photon. And so this would be our final answer to complete this example as our energy per photon for our orange light pending its wavelength. So if you have any questions on anything I reviewed, please leave them down below. Otherwise, I will see everyone in the next practice video.