Which of the following transitions in a hydrogen atom emits the photon of lowest frequency
a.n = 2 to n = 1 (B) n = 4 to n= 2 (C) n =4 to n =3 (D) n =3 to n = 1
please give solution?
2017-01-03 4:40 pm
回答 (1)
2017-01-03 5:02 pm
c : speed of light
ν : frequency of the photon emitted
λ : wavelength of the photon emitted
R : Rydberg constant
Rydberg formula : 1/λ = R [(1/n₁)² - (1/n₂)²]
Then, c/λ = cR [(1/n₁)² - (1/n₂)²]
It is known that : ν = c/λ
Hence, ν = cR [(1/n₁)² - (1/n₂)²]
(A) Frequency, ν = cR [(1/1)² - (1/2)²] = (3/4)cR = 108cR/144
(B) Frequency, ν = cR [(1/2)² - (1/4)²] = (3/16)cR = 27cR/144
(C) Frequency, ν = cR [(1/3)² - (1/4)²] = (7/144)cR = 7cR/144
(D) Frequency, ν = cR [(1/1)² - (1/3)²] = (8/9)cR = 128cR/144
Since 7cR/144 < 27cR/144 < 108cR/144 < 128cR/144
then the frequency in (C) is the lowest.
The answer: (C) n = 4 to n = 3
ν : frequency of the photon emitted
λ : wavelength of the photon emitted
R : Rydberg constant
Rydberg formula : 1/λ = R [(1/n₁)² - (1/n₂)²]
Then, c/λ = cR [(1/n₁)² - (1/n₂)²]
It is known that : ν = c/λ
Hence, ν = cR [(1/n₁)² - (1/n₂)²]
(A) Frequency, ν = cR [(1/1)² - (1/2)²] = (3/4)cR = 108cR/144
(B) Frequency, ν = cR [(1/2)² - (1/4)²] = (3/16)cR = 27cR/144
(C) Frequency, ν = cR [(1/3)² - (1/4)²] = (7/144)cR = 7cR/144
(D) Frequency, ν = cR [(1/1)² - (1/3)²] = (8/9)cR = 128cR/144
Since 7cR/144 < 27cR/144 < 108cR/144 < 128cR/144
then the frequency in (C) is the lowest.
The answer: (C) n = 4 to n = 3
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