The excitation of an electron on the surface of a photocell required [tex]5.0 \times 10^{-27} \text{ J}[/tex] of energy. Calculate the wavelength of light that was needed to excite the electron in an atom on the photocell.
Answer (2)
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.)
The wavelength of light needed to excite the electron in the photocell is approximately 39.78 meters, calculated using the formula Wavelength = E h ⋅ c . This wavelength corresponds to extremely low-frequency radio waves. Such a lengthy wavelength indicates a surprising response range for the photocell.
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