Answer:



In a hydrogen atom, an electron is in the nth excited state. It may come down to the 2nd excited state by emitting ten different wavelengths. We need to determine the value of n.



We know that the energy of an electron in the nth excited state of a hydrogen atom is given by the formula:

E_n = -13.6/n² eV

When an electron moves from a higher energy state to a lower energy state, it releases energy in the form of electromagnetic radiation. The energy of the emitted radiation is given by the formula:

ΔE = E_i - E_f = hf

where ΔE is the energy of the emitted radiation, E_i is the initial energy level, E_f is the final energy level, h is Planck's constant, and f is the frequency of the radiation.

Since the electron in the hydrogen atom moves from the nth excited state to the 2nd excited state, we can write:

ΔE = E_n - E₂ = hf

Substituting the values of the energies, we get:

-13.6/n² - (-13.6/2²) = hf

Simplifying, we get:

13.6(1/4 - 1/n²) = hf

We know that the frequency of the emitted radiation is related to the wavelength by the formula:

c = λf

where c is the speed of light, λ is the wavelength, and f is the frequency.

Since we are given ten different wavelengths, we can write:

λ₁, λ₂, λ₃, ..., λ₁₀

Substituting the value of f in terms of λ, we get:

c/λ₁ = 13.6(1/4 - 1/n²)
c/λ₂ = 13.6(1/4 - 1/n²)
c/λ₃ = 13.6(1/4 - 1/n²)
...
c/λ₁₀ = 13.6(1/4 - 1/n²)

We can simplify this set of equations by dividing each equation by the other to eliminate the constant 13.6:

λ₁/λ₂ = λ₃/λ₂ = ... = λ₁₀/λ₂

This means that all the wavelengths are related to each other by the same ratio. We can write:

λ₁ = kλ₂
λ₃ = kλ₂
...
λ₁₀

Answer is 6th excited state