Question #0a254
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Please see the explanation.
The wavelength of #550" kHz"# is:
#(3.0xx10^8" m/s")/(5.5xx10^5" Hz")= 545.5" m"#
The wavelength of #1600" kHz"# is:
#(3.0xx10^8" m/s")/(1.6xx10^6" Hz")= 187.5" m"#
The wavelength of 88 MHz is:
#(3.0xx10^8" m/s")/(8.8xx10^7" Hz")= 3.4" m"#
The wavelength of 108 MHz is:
#(3.0xx10^8" m/s")/(1.08xx10^8" Hz")= 2.8" m"#
AM signals: #188"m"-545"m"#
FM signals: #2.8"m"-3.4"m"#
From the chemistry reference sheet, #C=lambdanu# where #C# is the speed of the wave, #lambda# is the wavelength, and #nu# is the frequency. Solving this formula for wavelength, #lambda=C/nu#.
For AM signals, #C=3.0times10^8"m/s"# and #nu# ranges from #550times10^3"Hz"# to #1600times10^3"Hz"# (since #1"kHz"=10^3"Hz"#). Therefore, the wavelength ranges from #lambda=(3.0times10^8"m/s")/(550times10^3"Hz")=545"m"# to #lambda=(3.0times10^8"m/s")/(1600times10^3"Hz")=188"m"#.
For FM signals, #C=3.0times10^8"m/s"# and #nu# ranges from #88times10^6"Hz"# to #108times10^6"Hz"# (since #1"MHz"=10^6"Hz"#). Therefore, the wavelength ranges from #lambda=(3.0times10^8"m/s")/(88times10^6"Hz")=3.4"m"# to #lambda=(3.0times10^8"m/s")/(108times10^6"Hz")=2.8"m"#.
