To find the frequency and energy of an X-ray with a given wavelength of [tex]\(2.090 \times 10^{-11} \)[/tex] meters, we can use the following formulae:
1. Calculating Frequency:
The relationship between the speed of light [tex]\(c\)[/tex], wavelength [tex]\(\lambda\)[/tex], and frequency [tex]\(f\)[/tex] is given by the formula:
[tex]\[
f = \frac{c}{\lambda}
\][/tex]
Where:
- [tex]\( c \)[/tex] is the speed of light in a vacuum, approximately [tex]\( 3 \times 10^8 \)[/tex] meters per second.
- [tex]\( \lambda \)[/tex] is the wavelength of the X-ray, given as [tex]\( 2.090 \times 10^{-11} \)[/tex] meters.
Substituting the given values:
[tex]\[
f = \frac{3 \times 10^8}{2.090 \times 10^{-11}}
\][/tex]
Performing the division:
[tex]\[
f \approx 1.435 \times 10^{19} \text{ Hz}
\][/tex]
Therefore, the frequency of the X-ray is approximately [tex]\( 1.435 \times 10^{19} \)[/tex] Hertz.
2. Calculating Energy:
The energy [tex]\(E\)[/tex] of a photon is given by Planck's equation:
[tex]\[
E = h \cdot f
\][/tex]
Where:
- [tex]\( h \)[/tex] is Planck's constant, approximately [tex]\( 6.62607015 \times 10^{-34} \)[/tex] Joule seconds (Js).
- [tex]\( f \)[/tex] is the frequency which we have calculated as [tex]\( 1.435 \times 10^{19} \)[/tex] Hz.
Substituting the values:
[tex]\[
E = 6.62607015 \times 10^{-34} \times 1.435 \times 10^{19}
\][/tex]
Performing the multiplication:
[tex]\[
E \approx 9.511 \times 10^{-15} \text{ Joules}
\][/tex]
Thus, the energy of the X-ray photon is approximately [tex]\( 9.511 \times 10^{-15} \)[/tex] Joules.
To summarize, for an X-ray with a wavelength of [tex]\( 2.090 \times 10^{-11} \)[/tex] meters:
- The frequency is approximately [tex]\( 1.435 \times 10^{19} \)[/tex] Hertz.
- The energy is approximately [tex]\( 9.511 \times 10^{-15} \)[/tex] Joules.
