Jika 5x= 60, maka nilai x adalah
Answer (4)
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.
;
Jawaban:12Penjelasan dengan langkah-langkah:( . ) adalah perkalian dalam mtk5x = 605 . x = 60 karena dikali yang sama" tinggal di pindah ruas saja menjadi pembagian x = 60 ÷ 5x = 12
•°• Jika 5x = 60, maka nilai x adalah 12.[tex] \\ \\ [/tex]Penjelasan dengan langkah-langkah:5x = 60x = 60/5x = 12[tex]~[/tex][tex]__________________________________________________________________________________________[/tex][tex] \\ \\ [/tex] [tex]\blue{\boxed{\colorbox{skyblue}{\rm{- AvR}}}}[/tex]
