What is the longest wavelength of electromagnetic radiation capable of ionizing this hydrogen atom in an excited state? Report your answer in nanometers.

Question

What is the longest wavelength of electromagnetic radiation capable of ionizing this hydrogen atom in an excited state? Report your answer in nanometers.

Δ𝐸=−2.178×10−18 J(1𝑛2final−1𝑛2initial)

Answer 1

Calculate dE, then dE = hc/wavelength. Solve for wavelength.

Answer 2

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:

Δ𝐸 = -2.178×10^(-18) J(1/𝑛2final - 1/𝑛2initial)

In this formula, 𝜆 represents the wavelength of the electromagnetic radiation, 𝑛final is the final energy level of the hydrogen atom, and 𝑛initial is the initial energy level.

For ionization to occur, the final energy level (𝑛final) 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:

Δ𝐸 = -2.178×10^(-18) J(1/∞ - 1/𝑛2initial)

As 1/∞ approaches zero, the equation simplifies to:

Δ𝐸 ≈ -2.178×10^(-18) J(0 - 1/𝑛2initial)

Now, to find the maximum wavelength (λ), we can use the relationship between energy and wavelength:

𝐸 = 𝑛𝑢𝑚𝑒𝑟𝑖𝑐⎛⎝ℎ𝑐𝜆⎞⎠

Where:
𝐸 is the energy in joules
ℎ is Planck's constant (6.62607015 × 10^(-34) J·s)
𝑐 is the speed of light (2.998 × 10^(8) m/s)
𝜆 is the wavelength in meters

Rearranging the formula, we get:

𝜆 = 𝑐ℎ / 𝐸

Substituting the value of Δ𝐸 into the equation, we have:

𝜆 = 𝑐ℎ / (-2.178×10^(-18) J(0 - 1/𝑛2initial))

Now, we can convert the wavelength from meters to nanometers by multiplying by 10^9 (since 1 meter = 10^9 nanometers):

𝜆 (in nanometers) = (𝑐ℎ / (-2.178×10^(-18) J(0 - 1/𝑛2initial))) * 10^9

To calculate the answer, you'll need to provide the value of 𝑛2initial in the formula.