Use hc/lambda = E [energy]
So lambda = hc / E
Use h = 6.626 x 10^-34 and c = 3 x 10^8
Ofcourse E is given in joule
You could solve by yourself
-- The question doesn't specify the type of device where this cathode is found.I have to assume ... since a range is specified for the assumed wavelength ofvisible light, and because the question is posted to the < Visible Light > category ...that we're dealing here with a device designed to detect visible light, such as aphotodiode.-- The photons with the greatest energy will be those corresponding to the highestfrequency/shortest wavelength ... 390 nm .-- 1 Joule = 6.24 x 1018 eV-- So the cathode work function must behc/λ = (6.63 x 10-34) x (3 x 108) x (6.24 x 1018)/390 x 10-9 = 3.18 eVor less, in order for the circuit to respond to visible light.This number sounds quite reasonable for the process we're discussing here,so I'm just going to leave it like that and quietly tip-toe out of the room.
A typical human eye will respond to wavelengths from about 390 to 750 nanometers. (0.00039 to 0.00075 millimeter)
For a normal human being, visible light comprises electromagnetic radiation with wavelength in the range in the 390 nanometres (frequency = 770 THz) to 700 nm (430 THz).
390 grams = 13.76 ounces
390 meters = 1,279.53 feet.
Electromagnetic radiation between 490-790 terahertz, or about 390-700 nanometers wavelength.
-- The question doesn't specify the type of device where this cathode is found.I have to assume ... since a range is specified for the assumed wavelength ofvisible light, and because the question is posted to the < Visible Light > category ...that we're dealing here with a device designed to detect visible light, such as aphotodiode.-- The photons with the greatest energy will be those corresponding to the highestfrequency/shortest wavelength ... 390 nm .-- 1 Joule = 6.24 x 1018 eV-- So the cathode work function must behc/λ = (6.63 x 10-34) x (3 x 108) x (6.24 x 1018)/390 x 10-9 = 3.18 eVor less, in order for the circuit to respond to visible light.This number sounds quite reasonable for the process we're discussing here,so I'm just going to leave it like that and quietly tip-toe out of the room.
Ultra-violet light (wavelength approx 10-390 nanometers) lies between visible light (wavelength approx 390-750 nanometers) and X-rays (wavelength approx 0.1-10 nanometers).
A monochromatic source is a source of light of a discrete wavelength. White light is a mixture photons with wavlengths from 390 to 750 nm (what the human eye can detect). The monochrmatic light will have a specific wavelength. For example all photons have wavelength 200 nm.
A nanometre (nm). Human eyes are sensitive to light in the ranges from 390 nm to 700 nm.
A typical human eye will respond to wavelengths from about 390 to 750 nanometers. (0.00039 to 0.00075 millimeter)
If it is between 390 and 700nm (nano meter) then yes.
Any electromagnetic energy whose wavelength is between roughly 390 and 750 nanometers is visible, if some of it happens to enter your eye. By traditional usage as well as international agreement, this type of electromagnetic energy is widely referred to as "visible light".
390 squared is 390 times 390 = 152,100
46% of 390= 46% * 390= 0.46 * 390= 179.4
390
390 is divisible by 3: 390/3 = 130 390 is not divisible by 9: 390/9 = 43.3 recurring (that is, 43.3333..) 390 is divisible by 10: 390/10 = 39