Question

What inductance must be connected to a$\mathbf{17}\mathbf{}\mathbf{pF}$capacitor in an oscillator capable of generating$\mathbf{550}\mathbf{}\mathbf{nm}$(i.e., visible) electromagnetic waves? Comment on your answer.

Short answer

The inductance of the oscillator is .$5.00\times {10}^{-21}\text{H}$

Step-by-step solution

Write the given quantities

Capacitor,

$\begin{array}{rcl}C& =& 17\text{pf}\\ & =& 17\times {10}^{-12}\text{f}\end{array}$

Wavelength,

$\begin{array}{rcl}\lambda & =& 550\text{nm}\\ & =& 550\times {10}^{-9}\text{m}\end{array}$

Velocity of light

$c=3\times {10}^8\frac{\text{m}}{\text{s}}$

Determine the concepts of inductance and LC circuit

If f is the frequency and𝛌is the wavelength of an electromagnetic wave in the LC circuit of a generator, the value for inductance L is found using the formula for frequency of an LC circuit.

Formula:

$f=\frac{1}{2\pi \sqrt{LC}}$

$f=\frac{c}{\lambda }$

Determine the inductance of the oscillator

Consider the formula for the frequency as:

$f=\frac{1}{2\pi \sqrt{LC}}$

But we know that

$f=\frac{c}{\lambda }$

Therefore,
$\frac{c}{\lambda }=\frac{1}{2\pi \sqrt{LC}}$

Solving for L,

$L=\frac{{\lambda }^2}{4{\pi }^{2}C{c}^2}$

Substitute the values and solve as:

$\begin{array}{rcl}L& =& \frac{{\left(550\times {10}^{-9}\right)}^2}{4\times {\left(3.14\right)}^2\times \left(17\times {10}^{-12}\right)\times {\left(3\times {10}^8\right)}^2}\\ & =& 5.00\times {10}^{-21}\text{H}\end{array}$

The inductance of the oscillator is $5.00\times {10}^{-21}\mathrm{H}$.