The excitation of an electron on the surface of a photocell required 5.0 x 10 –27 J of energy. Calculate the wavelength of light that was needed to excite the electron in an atom on the photocell

c = speed of light in vacuum = about 3 x 10⁸ meters/second  h = Planck's Konstant = 6.63 x 10⁻ ³⁴ joule-secondEnergy = (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 metersThis is an astonishing result !  Simply amazing.  That wavelength correspondsto a frequency of about 7.54 MHz, in one of the short-wave radio bands used bya 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.)