The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot?

We are told that the modulation frequency of the existing amplitude-modulated (AM) station is $$f_m = 250\ \text{kHz}$$ and that this value is 10 % of the carrier frequency. Mathematically, the statement “250 kHz is 10 % of the carrier” translates to

$$f_m \;=\; 0.10\,f_c$$

Substituting the given numerical value of $$f_m$$, we have

$$250\ \text{kHz} \;=\; 0.10\,f_c$$

To isolate $$f_c$$ we divide both sides by $$0.10$$:

$$f_c \;=\; \frac{250\ \text{kHz}}{0.10}$$

Carrying out the division,

$$f_c \;=\; 2500\ \text{kHz}$$

In AM transmission the occupied spectrum stretches from the lower side-band to the upper side-band. First we recall the definitions:

Upper side-band frequency:

$$f_{\text{USB}} \;=\; f_c + f_m$$

Lower side-band frequency:

$$f_{\text{LSB}} \;=\; f_c - f_m$$

Now we substitute $$f_c = 2500\ \text{kHz}$$ and $$f_m = 250\ \text{kHz}$$ into these formulas.

For the upper side-band:

$$f_{\text{USB}} \;=\; 2500\ \text{kHz} \;+\; 250\ \text{kHz} \;=\; 2750\ \text{kHz}$$

For the lower side-band:

$$f_{\text{LSB}} \;=\; 2500\ \text{kHz} \;-\; 250\ \text{kHz} \;=\; 2250\ \text{kHz}$$

Hence the complete spectrum occupied by the existing station extends from $$2250\ \text{kHz}$$ up to $$2750\ \text{kHz}$$. Any second station must be assigned a carrier frequency that lies outside this interval so that its own side-bands will not overlap the first station’s band.

Examining the four options:

• $$2750\ \text{kHz}$$ falls exactly on the upper side-band edge, so it would interfere.

• $$2250\ \text{kHz}$$ falls exactly on the lower side-band edge, so it would also interfere.

• $$2900\ \text{kHz}$$ is above the upper edge by only $$150\ \text{kHz}$$, which is less than the required guard distance of one modulation frequency (250 kHz); its lower side-band would still overlap.

• $$2000\ \text{kHz}$$ is $$250\ \text{kHz}$$ below the lower edge, so even its upper side-band $$\bigl(2000\ \text{kHz} + 250\ \text{kHz} = 2250\ \text{kHz}\bigr)$$ just touches but does not overlap the existing band, providing the needed separation.

Therefore the safest and correct choice for the new broadcast frequency is

$$f_{\text{new}} = 2000\ \text{kHz}$$

Hence, the correct answer is Option 2.