A hypothetical electromagnetic wave is show below. 
Correct Answer: 5
Approach Solution - 1
The frequency \( f \) of an electromagnetic wave is related to its wavelength \( \lambda \) by the equation \( f = \frac{c}{\lambda} \), where \( c \) is the speed of light (~\( 3 \times 10^8 \) m/s). Given \( \lambda = 1.5 \) pm, we convert it to meters: \( 1.5 \) pm = \( 1.5 \times 10^{-12} \) m.
Now, substitute the values:
\( f = \frac{3 \times 10^8}{1.5 \times 10^{-12}} \approx 2 \times 10^{20} \) Hz.
This corresponds to \( x = 20 \).
Given range: 5 to 5 (interpreted as 5,5 meaning \( x \) must be an integer between 5 and 5). However, this seems inconsistent but if taken as a possible typo, since we calculated \( x \) as \( 20 \), let's verify:
\( x = 20 \) does not fit within 5. Hence, verification may be needed on range specification.
The nearest integer value of \( x \) remains \( 20 \).
Approach Solution -2
Given:
\[\lambda = 1.5 \times 4 \, \text{pm} = 6 \times 10^{-12} \, \text{meter}.\]
Using the relationship:
\[\lambda \nu = c,\]
where $c = 3 \times 10^8 \, \text{m/s}$, we can find $\nu$ as:
\[6 \times 10^{-12} \cdot \nu = 3 \times 10^8\]
\[\nu = \frac{3 \times 10^8}{6 \times 10^{-12}} = 5 \times 10^{19} \, \text{Hz}.\]
Therefore, $x = 5$.
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