What is the longest wavelength of electromagnetic radiation capable of ionizing this hydrogen atom in an excited state? Report your answer in nanometers.
ΞπΈ=β2.178Γ10β18 J(1π2finalβ1π2initial)
Calculate dE, then dE = hc/wavelength. Solve for wavelength.
To find the longest wavelength of electromagnetic radiation capable of ionizing a hydrogen atom in an excited state, we need to use the formula you have provided:
ΞπΈ = -2.178Γ10^(-18) J(1/π2final - 1/π2initial)
In this formula, π represents the wavelength of the electromagnetic radiation, πfinal is the final energy level of the hydrogen atom, and πinitial is the initial energy level.
For ionization to occur, the final energy level (πfinal) should be infinity. This means that the hydrogen atom is completely separated into an electron and a proton. Therefore, we can rewrite the formula as:
ΞπΈ = -2.178Γ10^(-18) J(1/β - 1/π2initial)
As 1/β approaches zero, the equation simplifies to:
ΞπΈ β -2.178Γ10^(-18) J(0 - 1/π2initial)
Now, to find the maximum wavelength (Ξ»), we can use the relationship between energy and wavelength:
πΈ = ππ’πππππβββππββ
Where:
πΈ is the energy in joules
β is Planck's constant (6.62607015 Γ 10^(-34) JΒ·s)
π is the speed of light (2.998 Γ 10^(8) m/s)
π is the wavelength in meters
Rearranging the formula, we get:
π = πβ / πΈ
Substituting the value of ΞπΈ into the equation, we have:
π = πβ / (-2.178Γ10^(-18) J(0 - 1/π2initial))
Now, we can convert the wavelength from meters to nanometers by multiplying by 10^9 (since 1 meter = 10^9 nanometers):
π (in nanometers) = (πβ / (-2.178Γ10^(-18) J(0 - 1/π2initial))) * 10^9
To calculate the answer, you'll need to provide the value of π2initial in the formula.