Question

1. The ni = 2 to nf = 6 transition in the Bohr hydrogen atom corresponds to the a)_______ of a
photon with a wavelength of b)_______ nm.
a) Is this transition of energy state absorption of photon or emission of photon?
b) Calculate the corresponding wavelength(λ) of photon; nm = 10-9 m
(use Rydberg equation)
2. The lowest orbital energy is reached when the number of electrons with the same spin is
maximized. This statement describes ________________________.

Answer 1

Answer on Question #55760 - Chemistry - General chemistry

Question:

1. The ni = 2 to nf = 6 transition in the Bohr hydrogen atom corresponds to the a) ______ of a photon with a wavelength of b) ______ nm. a) Is this transition of energy state absorption of photon or emission of photon? b) Calculate the corresponding wavelength(λ) of photon; nm = 10⁻⁹ m (use Rydberg equation)². The lowest orbital energy is reached when the number of electrons with the same spin is maximized. This statement describes

Solution

1. This transition of energy state is absorption of photon because the transition proceeds for the lower to the upper energy level, so some amount of energy required to do that. The wavelength responsible for this transition can be calculated using Rydberg's equation:

$$\frac {1}{\lambda} = R _ {H} \left(\frac {1}{n _ {i} ^ {2}} - \frac {1}{n _ {f} ^ {2}}\right)$$ $$\frac {1}{\lambda} = 1.097 \times 10 ^ {7} \left(\frac {1}{2 ^ {2}} - \frac {1}{6 ^ {2}}\right) = 2.346 \times 10 ^ {6}$$ $$\lambda = \left(\frac {1}{2.346 \times 10 ^ {6}}\right) = 4.102 \times 10 ^ {- 7} \, \text{m or } 410 \, \text{nm}$$

The $n_i = 2$ to $n_f = 6$ transition in the Bohr hydrogen atom corresponds to the a) absorption of a photon with a wavelength of b) 410 nm.

2. The lowest orbital energy is reached when the number of electrons with the same spin is maximized. This statement describes Hund's rule.

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