Emission Spectral Lines & the Rydberg Formula

[ObliviousSage]

FIGURED IT OUT, PLEASE DISREGARD

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/$$\lambda$$=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?

 

[ObliviousSage]

This always happens, as soon as I finish typing up my question I figure out what I was doing wrong. >_<

I need n₂ as large as possible to get the smallest wavelength.