Question

CALCULATOR FULL SCREEN PRINTER VERSION BACK NEXT SPECTROSCOPY AND REACTIVITY T04/S07 The energy levels the electron can occupy in the Li2+ ion can be calculated using the energy level equation. A Li2+ ion emitted a photon with a frequency of 5.55 x 1015 Hz to reach an energy level with n=2. What was the value of n for its initial energy level? Speed of Light c=3.00x108 m/s Planck's Constant h=6.63x10-34 J s Rydberg's Constant Rh=2.18x10-18 ] Avogadro's No. NA=6.02x1023mol-1 Energy Level Eqn En(J)=-RH(22/n2) Z=nuclear charge n=quantum number exact number, no tolerance

Answer 1

Given data

Frequency = 5.55 x 10^15 Hz or s-1

Initial energy level = n1

Final energy level = n2 = 2

Wavelength = speed of light / frequency

= (3 x 10^8 m/s) / (5.55 x 10^15 s-1)

= 5.405 x 10^-8 m

Atomic number of Li, Z = 3

From the Rydberg's equation

1/wavelength = (RH x Z^2) [(1/n1^2) - (1/n2^2)]

1/(5.405 x 10^-8 m) = (1.097 x 10^7 m-1) x 3^2 [(1/n1^2) - (1/2^2)]

0.1873 = (1/n1^2) - 0.25

0.4373 = 1/n1^2

2.286 = n1^2

n1 = 1.51