the primary visible emissions from mercury are 404.7 nm and 435.8 nm. Calculate the frequencies for these emissions. Calculate the energy of a single photon and of a mole of photons of light with each of these wavelengths.
To calculate the frequency of an emission, you can use the equation:
f = c / λ
where f is the frequency, c is the speed of light (approximately 3.00 × 10^8 m/s), and λ is the wavelength in meters.
Let's calculate the frequencies for the given emissions:
For 404.7 nm:
λ = 404.7 nm = 404.7 × 10^(-9) m (converting from nm to meters)
f = c / λ = 3.00 × 10^8 m/s / (404.7 × 10^(-9) m)
f ≈ 7.41 × 10^14 Hz
For 435.8 nm:
λ = 435.8 nm = 435.8 × 10^(-9) m (converting from nm to meters)
f = c / λ = 3.00 × 10^8 m/s / (435.8 × 10^(-9) m)
f ≈ 6.88 × 10^14 Hz
Now, let's calculate the energy of a single photon using the formula:
E = hf
where E is the energy, h is Planck's constant (approximately 6.626 × 10^(-34) J⋅s), and f is the frequency.
For 404.7 nm:
E = (6.626 × 10^(-34) J⋅s) × (7.41 × 10^14 Hz)
E ≈ 4.90 × 10^(-19) J
For 435.8 nm:
E = (6.626 × 10^(-34) J⋅s) × (6.88 × 10^14 Hz)
E ≈ 4.54 × 10^(-19) J
To calculate the energy of a mole of photons, you can use Avogadro's number (approximately 6.022 × 10^23 mol^(-1)):
For 404.7 nm:
Energy of a mole of photons = (4.90 × 10^(-19) J) × (6.022 × 10^23 mol^(-1))
Energy of a mole of photons ≈ 2.95 × 10^5 J
For 435.8 nm:
Energy of a mole of photons = (4.54 × 10^(-19) J) × (6.022 × 10^23 mol^(-1))
Energy of a mole of photons ≈ 2.73 × 10^5 J
Please note that the values given here are approximations.
To calculate the frequency of an emission, we can use the formula:
f = c / λ
where f represents frequency, c is the speed of light (approximately 3.00 x 10^8 m/s), and λ is the wavelength in meters.
Let's start by calculating the frequency for the emission at 404.7 nm:
Given:
Wavelength (λ) = 404.7 nm = 404.7 x 10^(-9) m
Speed of light (c) = 3.00 x 10^8 m/s
Plugging these values into the formula, we have:
f = (3.00 x 10^8 m/s) / (404.7 x 10^(-9) m)
Simplifying the equation, we get:
f ≈ 7.42 x 10^14 Hz
So, the frequency of the emission at 404.7 nm is approximately 7.42 x 10^14 Hz.
Now let's calculate the frequency for the emission at 435.8 nm:
Given:
Wavelength (λ) = 435.8 nm = 435.8 x 10^(-9) m
Speed of light (c) = 3.00 x 10^8 m/s
Using the formula, we can calculate the frequency:
f = (3.00 x 10^8 m/s) / (435.8 x 10^(-9) m)
Simplifying the equation, we get:
f ≈ 6.88 x 10^14 Hz
Therefore, the frequency of the emission at 435.8 nm is approximately 6.88 x 10^14 Hz.
To calculate the energy of a single photon, we can use the formula:
E = hf
where E represents energy, h is Planck's constant (approximately 6.626 x 10^(-34) J·s), and f is the frequency in Hz.
For the emission at 404.7 nm:
Given:
Frequency (f) = 7.42 x 10^14 Hz
Planck's constant (h) = 6.626 x 10^(-34) J·s
Using the formula, we can calculate the energy of a single photon:
E = (6.626 x 10^(-34) J·s) x (7.42 x 10^14 Hz)
Simplifying the equation, we get:
E ≈ 4.91 x 10^(-19) J
Therefore, the energy of a single photon with a wavelength of 404.7 nm is approximately 4.91 x 10^(-19) Joules.
To calculate the energy of a mole of photons, we can use Avogadro's number, which is 6.022 x 10^23 photons per mole.
For the emission at 404.7 nm:
Energy of a mole of photons = (4.91 x 10^(-19) J) x (6.022 x 10^23)
Simplifying the equation, we get:
Energy of a mole of photons ≈ 2.96 x 10^5 J/mol
Therefore, the energy of a mole of photons with a wavelength of 404.7 nm is approximately 2.96 x 10^5 Joules/mol.
c = frequency*wavelength
404.7 and 435.8 are wavelenths in nm. Convert those to m.
E = hc/wavelength for one photon.
That x 6.022E23 for a mole of photons.