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ObliviousSage
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FIGURED IT OUT, PLEASE DISREGARD
Find the shortest wavelength of the Lyman series for a triply-ionized beryllium atom (Be3+). Express your answer in nanometers using four significant digits.
Rydberg's Formula: 1/[tex]\lambda[/tex]=RZ2(1/n12 - 1/n22)
For the Lyman series, n1 is 1.
Since we want the shortest wavelength, we want the smallest n2, which would be 2.
Beryllium's atomic number (Z) is 4.
Rydberg's Constant (R) is 1.097*107.
Thus 1/n12 is 1, and 1/n22 is 1/4, and their difference is 3/4.
3/4 * 1.097*107 * 42 = 131640000
The wavelength is one over that, or approximately 7.59647523549*10-9m.
In nanometers, to 4 significant digits, that's 7.596nm, yet the website insists that's incorrect (I also tried 7.597nm). What am I doing wrong?
Homework Statement
Find the shortest wavelength of the Lyman series for a triply-ionized beryllium atom (Be3+). Express your answer in nanometers using four significant digits.
Homework Equations
Rydberg's Formula: 1/[tex]\lambda[/tex]=RZ2(1/n12 - 1/n22)
The Attempt at a Solution
For the Lyman series, n1 is 1.
Since we want the shortest wavelength, we want the smallest n2, which would be 2.
Beryllium's atomic number (Z) is 4.
Rydberg's Constant (R) is 1.097*107.
Thus 1/n12 is 1, and 1/n22 is 1/4, and their difference is 3/4.
3/4 * 1.097*107 * 42 = 131640000
The wavelength is one over that, or approximately 7.59647523549*10-9m.
In nanometers, to 4 significant digits, that's 7.596nm, yet the website insists that's incorrect (I also tried 7.597nm). What am I doing wrong?