Answer : The wavelength of the signal is
. This is not a visible light.
Solution : Given,
Frequency = 927.9 MHz =
![(1MHz=10^6Hz)](https://img.qammunity.org/2019/formulas/chemistry/high-school/8ypcwsdmkfm7r90kxh7oe81e9u6ujchzbt.png)
Speed of light =
![3* 10^8m/s](https://img.qammunity.org/2019/formulas/chemistry/high-school/jga4ycqtgnkhppw5hldvght9qiskdu5yhi.png)
Formula used :
![\\u=(c)/(\lambda)\\\lambda=(c)/(\\u)](https://img.qammunity.org/2019/formulas/chemistry/middle-school/sklidzb8k0d6mjppezjs5mnr0b2y3st7s7.png)
where,
= frequency
= wavelength
c = speed of light
Now put all the given values in this formula, we get the wavelength of the signal.
![(1m=10^9nm)](https://img.qammunity.org/2019/formulas/chemistry/middle-school/ue130l770d53j5m34cxz2gqqpveumghjgg.png)
This is not a visible light because the range of visible light is (380 to 700 nm).
Therefore, the wavelength of the signal is
. This is not a visible light.