c = speed of light in vacuum = about 3 x 10⁸ meters/second
h = Planck's Konstant = 6.63 x 10⁻ ³⁴ joule-second
Energy = (h x frequency) = (h c / wavelength)
Wavelength = (h c) / (energy)
Wavelength = (6.63 x 10^-34 joule-sec x 3 x 10^8 meter/sec) / (5 x 10^-27 joule)
= 19.89 x 10^-26 / 5 x 10^-27 = 39.78 meters
This is an astonishing result ! Simply amazing. That wavelength corresponds
to a frequency of about 7.54 MHz, in one of the short-wave radio bands used by
a lot of foreign-broadcast stations.
If the number in the problem is correct, it means that this 'photocell' responds
to any electromagnetic signal at 7.54 MHz or above ... short-wave radio,
commercial FM or TV signals, FRS walkie-talkies, garage-door openers,
Bluetooth thingies, home WiFi boxes, WiFi from a laptop, microwave ovens,
cellphones, any signal from a satellite, any microwave dish, any heat lamp,
flashlight, LED, black light, or X-ray machine. Some "photocell" !
I'm thinking the number given in the problem for the energy of a photon
at the detection threshold of this device must be wrong by several orders
of magnitude.
(But my math is still bullet-proof.)
