A

Step 1: Understand the LCR Circuit Response
The impedance in a series LCR circuit is given by
$$ Z = R + j(\omega L - \frac{1}{\omega C}) $$
Where:
- $R$ is the resistance.
- $L$ is the inductance.
- $C$ is the capacitance.
- $\omega = 2\pi f$ is the angular frequency.

Step 2: Phase Angle and Current Amplitude
The phase angle $\phi$ between the voltage and current is given by
$$ \tan(\phi) = \frac{\omega L - \frac{1}{\omega C}}{R} $$
The current can be expressed as:
$$ I = \frac{V}{Z} $$
And at resonance, the current amplitude is maximum.

Step 3: Current Amplitude at Frequencies $f_1$ and $f_2$
Assuming current amplitude falls to $\frac{1}{\sqrt{2}}$ at frequencies $f_1$ and $f_2$, we use the relation:
$$ I(f) = I(\text{max}) \sqrt{1 - \frac{(f - f_0)^2}{\Delta f^2}} $$
Where $f_0$ is the resonant frequency and $\Delta f$ is the bandwidth.

Step 4: Bandwidth Calculation
The bandwidth $\Delta f$ is defined by
$$ \Delta f = \frac{R}{2\pi L} $$
Thus, the frequencies $f_1$ and $f_2$ can be expressed as:
$$ f_1 = f_0 - \frac{R}{4\pi L}, \quad f_2 = f_0 + \frac{R}{4\pi L} $$
From this, the separation is:
$$ f_2 - f_1 = \frac{R}{2\pi L} $$

Final Conclusion
Thus, the frequencies $f_1$ and $f_2$ at which the current amplitude falls to $\frac{1}{\sqrt{2}}$ are indeed separated by an interval equal to $\frac{R}{2\pi L}$. Therefore, the statement is proven.