Answer:
The shortest wavelength of an AM radio signal is 200 meters.
Explanation:
It is given that, AM radio signals broadcast electromagnetic radiation at frequencies between 590 kHz and 1.50 MHz.
[tex]f_1=590\ kHz=590\times 10^3\ Hz[/tex]
The speed of electromagnetic wave is given by :
[tex]c=f_1\lambda_1[/tex]
[tex]\lambda_1=\dfrac{c}{f_1}[/tex]
[tex]\lambda_1=\dfrac{3\times 10^8}{590\times 10^3}[/tex]
[tex]\lambda_1=508.47\ m[/tex]
If [tex]f_2=1.5\ MHz=1.5\times 10^6\ Hz[/tex]
[tex]c=f_2\lambda_2[/tex]
[tex]\lambda_2=\dfrac{c}{f_2}[/tex]
[tex]\lambda_2=\dfrac{3\times 10^8}{1.5\times 10^6}[/tex]
[tex]\lambda_2=200\ m[/tex]
So, the shortest wavelength of an AM radio signal is 200 meters. Hence, this is the required solution.