A photon has [tex]\(3.4 \times 10^{-18}\)[/tex] joules of energy. Planck's constant is [tex]\(6.63 \times 10^{-34} J \cdot s\)[/tex].

What is the frequency of the photon?

A. [tex]\(5.12 \times 10^{-15} Hz\)[/tex]
B. [tex]\(1.95 \times 10^{16} Hz\)[/tex]
C. [tex]\(5.12 \times 10^{15} Hz\)[/tex]
D. [tex]\(1.95 \times 10^{-16} Hz\)[/tex]



Answer :

To find the frequency of a photon given its energy, we can use the well-known relationship in quantum mechanics expressed by Planck's equation:

[tex]\[ E = h \cdot f \][/tex]

where:
- [tex]\( E \)[/tex] is the energy of the photon,
- [tex]\( h \)[/tex] is Planck's constant,
- [tex]\( f \)[/tex] is the frequency of the photon.

Given:
- [tex]\( E = 3.4 \times 10^{-18} \)[/tex] joules,
- [tex]\( h = 6.63 \times 10^{-34} \)[/tex] joules·second,

we need to find the frequency [tex]\( f \)[/tex]. Rearrange the equation to solve for [tex]\( f \)[/tex]:

[tex]\[ f = \frac{E}{h} \][/tex]

Substitute the given values into the equation:

[tex]\[ f = \frac{3.4 \times 10^{-18}}{6.63 \times 10^{-34}} \][/tex]

When you carry out this division, you get:

[tex]\[ f = 5.128205128205129 \times 10^{15} \][/tex]

This value can be approximated to scientific notation as:

[tex]\[ f \approx 5.12 \times 10^{15} \text{ Hz} \][/tex]

Thus, the frequency of the photon is [tex]\( 5.12 \times 10^{15} \)[/tex] Hz. Therefore, the correct answer is:

[tex]\[ \boxed{5.12 \times 10^{15} \text{ Hz}} \][/tex]