calculate the max wavelength of light capable of removing an electron for a hydrogen atom from the energy state characterized by the following :
n=2 in nm
n=4 in nm
E = 2.180 x 10^-18 J/n2 where n is the principle quantum number of 2 or 4 for your two problems.
Then E = hc/lambda. h is Planck's constant, c is speed of light, lambda is wavelength IN METERS. Solve for lambda.
To calculate the maximum wavelength of light capable of removing an electron from a hydrogen atom in a specific energy state, we can use the equation for the energy levels of the hydrogen atom:
E = -13.6 eV / n^2
where E is the energy in electron volts (eV) and n is the principal quantum number.
First, we need to find the energy difference between the two energy states. We subtract the energy of the initial state (n=2) from the energy of the final state (n=4):
ΔE = E_final - E_initial
= (-13.6 eV / 4^2) - (-13.6 eV / 2^2)
= -13.6 eV / 16 + 13.6 eV / 4
= -0.85 eV + 3.4 eV
= 2.55 eV
Next, we can calculate the maximum wavelength of light using the equation:
E = (hc) / λ
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the light.
First, convert the energy difference from electron volts (eV) to joules (J):
ΔE = 2.55 eV * (1.602 x 10^-19 J/eV)
= 4.09 x 10^-19 J
Rearrange the equation to solve for wavelength (λ):
λ = (hc) / ΔE
Plug in the values:
λ = (6.626 x 10^-34 J·s * 2.998 x 10^8 m/s) / (4.09 x 10^-19 J)
= 4.86 x 10^-7 m
Finally, convert the wavelength from meters (m) to nanometers (nm):
λ = 4.86 x 10^-7 m * (1 x 10^9 nm/m)
= 486 nm
Therefore, the maximum wavelength of light capable of removing an electron for the n=2 energy state of a hydrogen atom is approximately 486 nm.
Note: You can follow the same process to calculate the maximum wavelength for the n=4 energy state by substituting the appropriate values for E_initial and E_final in the above calculations.
To calculate the maximum wavelength of light capable of removing an electron from a specific energy state in a hydrogen atom, we will use the Rydberg formula. The Rydberg formula is given by:
1/λ = R * (1/n₁² - 1/n₂²)
Where:
λ is the wavelength of light in meters,
R is the Rydberg constant (approximately 1.097 × 10^7 m⁻¹),
n₁ is the principal quantum number of the lower energy state, and
n₂ is the principal quantum number of the higher energy state.
Let's calculate the maximum wavelength for the two given energy states:
1. For n₁ = 2:
Using the Rydberg formula, we have:
1/λ = R * (1/2² - 1/n₂²)
We need to find the maximum wavelength, so we assume that n₂ goes to infinity. This means:
1/n₂² ≈ 0
Therefore, the equation simplifies to:
1/λ ≈ R * (1/2² - 0)
1/λ ≈ R * (1/4)
Simplifying further:
1/λ ≈ R/4
To find the maximum wavelength, we can take the reciprocal of both sides:
λ ≈ 4/R
Substituting the value of R = 1.097 × 10^7 m⁻¹, we can calculate the approximate maximum wavelength as:
λ ≈ 4 / (1.097 × 10^7)
λ ≈ 3.64 × 10⁻⁷ meters
Converting this into nanometers:
λ ≈ 364 nm
Therefore, the maximum wavelength is approximately 364 nm for n = 2 in a hydrogen atom.
2. For n₁ = 4:
Using the same process as above, we can calculate the maximum wavelength:
1/λ ≈ R * (1/4² - 0)
1/λ ≈ R/16
λ ≈ 16/R
Substituting the value of R = 1.097 × 10^7 m⁻¹, we can calculate the approximate maximum wavelength as:
λ ≈ 16 / (1.097 × 10^7)
λ ≈ 1.46 × 10⁻⁶ meters
Converting this into nanometers:
λ ≈ 1460 nm
Therefore, the maximum wavelength is approximately 1460 nm for n = 4 in a hydrogen atom.