Answer:
a. 1875 nm
b. 4051 nm
c. 1282 nm
These all are infrared electromagnetic radiation.
Explanation:
Our strategy here is to utilize the Rydberg equation for hydrogen atom electronic transition.
1/λ = Rh x (1/n₁² - 1/n₂²) where λ is the wavelength
Rh is Rydberg constant
n₁ and n ₂ are the energy levels ( n₁ < n₂ )
Now lets star the calculations.
a. n₁ = 3, n₂ = 4
1/λ = 1.097 x 10⁷ /m x (1/3² - 1/4²) = 5.333 x 10⁵/m
λ = 1/(5.333 x 10⁵ /m) = 1.875 x 10⁻⁶ m
Converting λ to nanometers:
1.875 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1875 nm
b. n₁ = 4, n₂ = 5
1/λ = 1.097 x 10⁷ /m x (1/4² - 1/5²) = 2.468 x 10⁵/m
λ = 1/(2.468 x 10⁵/m) = 4.051 x 10⁻⁶ m
4.051 x 10⁻⁶ m x (1 x10⁹ nm/m) = 4051 nm
c. n₁ = 3, n₂ = 5
1/λ = 1.097 x 10⁷ /m x (1/3² - 1/5²) = 7801 x 10⁵/m
λ = 1/(7801 x 10⁵/m) = 1282 x 10⁻⁶ m
1282 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1282 nm
All of these transitions fall in the infrared region of the spectrum.
The wavelength of each transition is obtained from the Rydberg formula as shown.
Using the Rydberg formula;
1/λ = R(1/n2^2 - 1/n1^2)
Where;
λ = wavelength
R = Rydberg constant = 1.097 × 10^7 m-1
n1 = initial energy level
n2 = final energy level
For the transition n = 4 → n = 3
1/λ = 1.097 × 10^7 m-1(1/3^2 - 1/4^2)
λ = 1.879 × 10^-6 m or 1879 nm
For the transition n = 5 → n = 4
1/λ = 1.097 × 10^7 m-1(1/4^2 - 1/5^2)
λ = 4.051 × 10^-6 m = 4051 nm
For the transition n = 5 → n = 3
1/λ = 1.097 × 10^7 m-1(1/3^2 - 1/5^2)
λ = 1.284 × 10^-6
λ = 1284 nm
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