Question

an isolated atom of a certain element emits light of wavelength 605 nm when the atom falls from its fifth excited state into its second excited state. the atom emits a photon of wavelength 425 nm when it drops from its sixth excited state into its second excited state. find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state.

Answer 1

To find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state, use the energy difference between the two states and the ratio of energy differences for two given transitions.

Step-by-step explanation:

To find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state, we can use the energy difference between the two states. The energy difference between two states is inversely proportional to the wavelength of the emitted light. So, we can use the ratio of the energy differences for the two given transitions to find the wavelength of the desired transition.

Let's say the energy difference between the fifth and second excited states is represented as ΔE₁ and the energy difference between the sixth and second excited states is represented as ΔE₂. We can set up the following equation:

ΔE₁ / ΔE₂ = λ₁ / λ₂

Using the given wavelengths λ₁ = 605 nm and λ₂ = 425 nm, we can substitute these values into the equation and solve for the unknown wavelength:

ΔE₁ / ΔE₂ = 605 nm / 425 nm

After cross-multiplication and simplification, we get:

ΔE₁ = (605 nm / 425 nm) * ΔE₂

Now we can substitute the known values of ΔE₂ and solve for ΔE₁. Once we have the energy difference ΔE₁, we can use it to calculate the corresponding wavelength of the desired transition.