Final answer:
The required inductance to achieve a resonant frequency of 88.0 MHz with a 2.50 pF capacitor is 0.21 μH, corresponding to option (b). This is determined by rearranging the LC circuit resonant frequency formula and inserting the given values for frequency and capacitance.
Step-by-step explanation:
To achieve the resonant frequency of 88.0 MHz with a 2.50 pF capacitor, it is necessary to calculate the inductance (L) required using the formula for resonant frequency of an LC (inductor-capacitor) circuit:
f = 1 / (2π√(LC))
Where:
- f is the resonant frequency (88.0 MHz in this case),
- L is the inductance we need to find,
- C is the capacitance (2.50 pF), and
- π is Pi (≃ 3.14159).
Rearranging the formula to solve for L gives:
L = 1 / (4π²f²C)
Using the values:
- f = 88.0 x 10⁶ Hz,
- C = 2.50 x 10⁻¹² F.
Performing the calculation results in:
L = 1 / (4 x π² x (88.0 x 10⁶)² x (2.50 x 10⁻¹²))
L = 1 / (1.558 x 10¹)
L = 0.64 x 10⁻¶ H or 0.21 μH
Therefore, the required inductance to achieve a resonant frequency of 88.0 MHz with a 2.50 pF capacitor is 0.21 μH, which corresponds to option (b).