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The _______ of an atom are negatively charged.
2. Atoms & Elements
The Atom
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What electron transition in a hydrogen atom, starting from n = 8, will produce light of wavelength 389 nm?
A
n = 8 to n = 4
B
n = 8 to n = 2
C
n = 8 to n = 3
D
n = 8 to n = 5
Verified step by step guidance1
Identify the formula to use: The Rydberg formula for hydrogen atom transitions is \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( \lambda \) is the wavelength, \( R_H \) is the Rydberg constant (1.097 x 10^7 m^-1), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.
Convert the given wavelength from nanometers to meters: Since 1 nm = 1 x 10^-9 m, convert 389 nm to meters by multiplying by 1 x 10^-9.
Rearrange the Rydberg formula to solve for \( n_1 \): \( n_1 = \sqrt{\frac{1}{\left( \frac{1}{\lambda} \cdot R_H + \frac{1}{n_2^2} \right)}} \).
Substitute the known values into the equation: Use \( \lambda = 389 \times 10^{-9} \) m, \( R_H = 1.097 \times 10^7 \) m^-1, and \( n_2 = 8 \).
Calculate \( n_1 \) for each possible transition (n = 4, 3, 2, 5) and determine which value of \( n_1 \) results in a wavelength closest to 389 nm.
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