Answer to Question #244633 in Chemistry for NO NICKNAME

Question

Which of the following transitions (in a hydrogen atom) represents an emission of the longest wavelength photon?

a) n = 1 to n = 2

b) n = 4 to n = 2

c) n = 3 to n = 1

d) n = 5 to n = 4

e) n = 3 to n = 5

Answer 1

Solution:

Rydberg equation can be used:

1/λ = R_H × (1/n²_f - 1/n²_i)

where λ is the wavelength of the emitted photon, R_H is the Rydberg constant (R_H=1.097×10⁷ m⁻¹), and n_f and n_i are the final and initial energy states, respectively.

For an emission to occur, the hydrogen atom should move from a higher energy state to a lower energy state. In this case, choices (a) and (e) are incorrect since they indicate the excitation of the hydrogen atom from a lower to a higher energy level.

Thus:

(b) n_i = 4 to n_f = 2

1/λ = (1.097×10⁷ m⁻¹) × (1/2² - 1/4²) = 2056875

λ = 1 / 2056875 = 4.86×10⁻⁷ m⁻¹ = 486 nm

(c) n_i = 3 to n_f = 1

1/λ = (1.097×10⁷ m⁻¹) × (1/1² - 1/3²) = 9751111.111

λ = 1 / 9751111.111 = 1.0255×10⁻⁷ m⁻¹ = 102.55 nm

(d) n_i = 5 to n_f = 4

1/λ = (1.097×10⁷ m⁻¹) × (1/4² - 1/5²) = 246825

λ = 1 / 246825 = 405.1×10⁻⁶ m⁻¹ = 4051 nm

102.55 nm (c) < 486 nm (b) < 4051 nm (d)

Based on the obtained values, the transition that represents the emission of the longest wavelength photon is n = 5 to n = 4 (i.e., 4050 nm).

Answer: d) n = 5 to n = 4