The first ionization energy of a carbon atom is 1.81 aJ. What is the frequency and wavelength, in nanometers, of photons capable of just ionizing carbon atoms?
v=______s^(-1)
lambda =_____nm
Assuming an ionization efficiency of 47.0%, how many such photons are needed to ionize 1.00 × 1016 atoms?
=_______protons
b,b.m
To find the frequency (v) and wavelength (lambda) of photons capable of just ionizing carbon atoms, we can use the equation:
c = λv
where c is the speed of light in a vacuum (approximately 3.00 × 10^8 m/s), λ is the wavelength, and v is the frequency.
First, let's find the frequency (v):
v = 1.81 aJ / h
Note that the ionization energy is given in attojoules (aJ) and we need to convert it to joules (J). 1 aJ is equal to 10^(-18) J. Here, h is the Planck's constant (approximately 6.626 × 10^(-34) J·s).
v = (1.81 × 10^(-18) J) / (6.626 × 10^(-34) J·s)
Simplifying,
v = 2.74 × 10^15 s^(-1)
Next, let's find the wavelength (lambda):
c = λv
λ = c / v
Remember, we need to convert the speed of light (c) to nanometers (nm). 1 nanometer is equal to 10^(-9) meters.
λ = (3.00 × 10^8 m/s) / (2.74 × 10^15 s^(-1))
λ = (3.00 × 10^8 m/s) × (1 nm / (10^(-9) m)) / (2.74 × 10^15 s^(-1))
Simplifying,
λ = 109.5 nm
So, the frequency (v) is approximately 2.74 × 10^15 s^(-1), and the wavelength (lambda) is approximately 109.5 nm.
Now, let's calculate the number of photons needed to ionize 1.00 × 10^16 carbon atoms, assuming an ionization efficiency of 47.0%.
The number of photons (N) can be found using the equation:
N = (n / ε) × (N_A / 100)
where n is the number of atoms to be ionized, ε is the ionization efficiency (in decimal form), N_A is Avogadro's number (approximately 6.022 × 10^23 particles/mol), and 100 is used for percentage conversion.
N = (1.00 × 10^16 atoms / 0.47) × (6.022 × 10^23 particles/mol / 100)
Simplifying,
N = 1.28 × 10^24 photons
So, to ionize 1.00 × 10^16 carbon atoms with an ionization efficiency of 47.0%, approximately 1.28 × 10^24 photons are needed.
Lastly, there is no information provided to calculate the number of protons, so that part of the question cannot be answered.