To find the energy of a photon given its wavelength, you can use the formula:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of the photon
- [tex]\( h \)[/tex] is Planck's constant ([tex]\(6.626 \times 10^{-34} \)[/tex] Joule seconds)
- [tex]\( c \)[/tex] is the speed of light ([tex]\(2.998 \times 10^8 \)[/tex] meters per second)
- [tex]\( \lambda \)[/tex] is the wavelength of the photon
In this problem, the wavelength ([tex]\( \lambda \)[/tex]) is given as 448 nanometers (nm).
Firstly, we need to convert the wavelength from nanometers to meters because the constants [tex]\( h \)[/tex] and [tex]\( c \)[/tex] are in SI units (meters, seconds, Joules):
1 nanometer (nm) = [tex]\( 1 \times 10^{-9} \)[/tex] meters (m).
Thus:
[tex]\[ \lambda = 448 \, \text{nm} = 448 \times 10^{-9} \, \text{m} \][/tex]
Now, substituting the values of [tex]\( h \)[/tex], [tex]\( c \)[/tex], and [tex]\( \lambda \)[/tex] into the equation, we get:
[tex]\[ E = \frac{6.626 \times 10^{-34} \, \text{Joule} \cdot \text{seconds} \times 2.998 \times 10^8 \, \text{meters/second}}{448 \times 10^{-9} \, \text{meters}} \][/tex]
By performing the calculation, the energy [tex]\( E \)[/tex] of the photon is:
[tex]\[ E \approx 4.434095535714285 \times 10^{-19} \, \text{J} \][/tex]
From the given choices:
A. [tex]\(8.91 \times 10^{-23} J\)[/tex]
B. [tex]\(2.97 \times 10^{-22} J\)[/tex]
C. [tex]\(4.44 \times 10^{-19} J\)[/tex]
D. [tex]\(1.48 \times 10^{-27} J\)[/tex]
The correct answer is:
C. [tex]\(4.44 \times 10^{-19} J\)[/tex]
