A monochromatic light source used in a diffraction experiment has a wavelength of 4.60 x 10-
7 m. (a) What is the frequency of this light? When you input large numbers such as those in
this problem, then input the numbers with parenthesis so as to obtain the correct answer. (b)
What is the energy of a photon of this light?
f = speed/wavelength = 3*10^8m/s / 4.6*10^-7m = 6.5*10^14/s
E = h*f = 6.6*10^-34 Js * 6.5*10^14/s = 4.3*10^-19J
Epic
(a) Ah, the wavelength of 4.60 x 10^-7 m, huh? Well, to find the frequency of this light, I hope you're ready for some math gymnastics! The frequency (f) is related to the wavelength (λ) through the equation v = c/λ, where v is the frequency and c is the speed of light. Since c is approximately 3.0 x 10^8 m/s, we can plug in the values to get the frequency. So, (drumroll please)... the frequency is (3.0 x 10^8 m/s) / (4.60 x 10^-7 m). Ta-da!
(b) Now, for the energy of a photon of this light, we can use another equation. The energy (E) of a photon is given by the equation E = hf, where h is Planck's constant (approximately 6.63 x 10^-34 J*s) and f is the frequency. So let's do some more calculations: (6.63 x 10^-34 J*s) * (frequency from part a). And voilà! You've got the energy of a photon of this light.
Remember, this is all assuming you wanted an actual answer and not just a bunch of clown nonsense. But hey, clown nonsense is more fun, right?
To calculate the frequency of the light, we can use the following formula:
\( c = \lambda \cdot \nu \)
Where:
- c is the speed of light (approximately \( 3.00 \times 10^8 \, \text{m/s} \))
- \( \lambda \) is the wavelength of the light
- \( \nu \) is the frequency of the light
Substituting the given values, we can solve for the frequency:
\( \nu = \frac{c}{\lambda} \)
\( \nu = \frac{3.00 \times 10^8 \, \text{m/s}}{4.60 \times 10^{-7} \, \text{m}} \)
Calculating the above expression:
\( \nu = 6.52 \times 10^{14} \, \text{Hz} \)
Therefore, the frequency of the light is \( 6.52 \times 10^{14} \, \text{Hz} \).
To calculate the energy of a photon of this light, we can use the formula:
\( E = h \cdot \nu \)
Where:
- E is the energy of the photon
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{J \cdot s} \))
- \( \nu \) is the frequency of the light
Substituting the given values, we can solve for the energy:
\( E = (6.63 \times 10^{-34} \, \text{J \cdot s}) \cdot (6.52 \times 10^{14} \, \text{Hz}) \)
Calculating the above expression:
\( E \approx 4.32 \times 10^{-19} \, \text{J} \)
Therefore, the energy of a photon of this light is approximately \( 4.32 \times 10^{-19} \, \text{J} \).
To find the frequency of light with a given wavelength, we can use the formula:
frequency = speed of light / wavelength
The speed of light, denoted by "c," is approximately 3.00 x 10^8 m/s.
(a) Therefore, to find the frequency of light with a wavelength of 4.60 x 10^-7 m, we can substitute these values into the formula:
frequency = (3.00 x 10^8 m/s) / (4.60 x 10^-7 m)
Calculating this expression will give us the frequency of the light.
(b) To find the energy of a photon of this light, we can use the equation:
energy = Planck's constant * frequency
Planck's constant, denoted by "h," is approximately 6.63 x 10^-34 J·s.
Substituting the values we obtained in part (a) for the frequency, we can calculate the energy of a photon of this light.