An electron in the n = 5 excited state of a hydrogen atom emits a photon of 1281 nm light. To what energy level does the electron move?
Any ideas on an equation for this? Is it some kind of backward version of Rydburg Eq?
conver nm to energy.
E=h*c/lambda
then, having E, find the n.
h*c/lambda
would this be planck's constant times speed of light / wavelength?
Yes. Another way to do it
(1/wavelength) = R(1/n^2 - 1/5^2)
where R is the Rydberg constant. Solve for n.
How many photons of light are emitted when the electrons in 10 hydrogen atoms drop from energy level 5 to 3?
Yes, you're on the right track! To find the energy level to which the electron moves, we need to use the Rydberg equation in a slightly modified form.
The Rydberg equation is usually used to calculate the wavelength of light emitted or absorbed when an electron transitions between energy levels in a hydrogen atom. However, we can rearrange the equation to solve for the energy level.
The general form of the Rydberg equation is:
1/λ = R * (1/n₁² - 1/n₂²)
where λ is the wavelength of the emitted or absorbed light, R is the Rydberg constant, and n₁ and n₂ represent the initial and final energy levels of the electron.
To solve for the final energy level (n₂), we can rearrange the equation as follows:
1/n₂² = 1/n₁² - 1/λR
In this case, we know that the initial energy level (n₁) is 5 and the wavelength (λ) of the emitted light is 1281 nm (or 1.281 μm). The Rydberg constant (R) can be approximated as 1.097 × 10⁷ m⁻¹.
Plugging these values into the equation, we can calculate the final energy level (n₂) to which the electron moves.
1/n₂² = 1/5² - 1/(1.281 × 10⁻⁶ * 1.097 × 10⁷)
Simplifying this expression will give us the reciprocal of n₂ squared. From there, we can solve for n₂ by taking the reciprocal of the square root.
Let's do the calculations to find the value of n₂.