Respuesta :
Answer: a) The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.
b) The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.
Explanation:
The formula that relates wavelength of emissions to Rydberg's constant and the n₁ values is
(1/λ) = R ((1/(n₁^2)) - (1/(n2^2))
Where λ = wavelength, R = (10.972 × 10^6)/m, n2 = ∞ (since they're emitted out of the atom already)
a) n₁ = ?
λ = 3282 nm = (3.282 × 10^-6)m
(1/(3.282 × 10^-6)) = (10.972 × 10^6) ((1/(n₁^2) (since 1/∞ = 0)
n₁^2 = (3.282 × 10^-6) × (10.972 × 10^6) = 36
n₁ = 6.
The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.
b) n₁ = ?
λ = 7460 nm = (7.46 × 10^-6)m
(1/(7.46 × 10^-6)) = (10.972 × 10^6) ((1/(n₁^2)) - (1/(n2^2)) for lowest energy line, n2 = n₁ + 1
(n₁^2)((n₁+1)^2))/(2n₁+1) = (7.46 × 10^-6) × (10.972 × 10^6) = 81.85
(n₁^2)((n₁+1)^2))/(2n₁+1) = 81.85
Solving the quadratic eqn,
n₁ = 5.
The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.
QED!
