Answer:
The answer is "[tex]3.83 \times 10^9 \ m[/tex]"
Explanation:
Z=2, so the equation is [tex]E= \frac{-4B}{n^2}[/tex]
Calculate the value for E when: Â
n=2 and n=9
The energy is the difference in transformation, name the energy delta E Deduct these two energies Â
In this transition, the wavelength of the photon emitted is:
[tex]\Delta E=2.18 \times 10^{-18} ( \frac{1}{4}- \frac{1}{81})[/tex]
[tex]\lambda = \frac{h c}{\Delta E}[/tex]
[tex]h ( Planck's\ constant) = 6.62 \times 10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times 10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\[/tex]
