Question

Advance Study Assignment: The Atomic Spectrum of Hydrogen 1. Found in the gas phase, the beryllium trication (triply charged positive, or +3, ion). Bet, has an energy level formula analogous to that of the hydrogen atom, since both species have only one electron. The en- ergy levels of the Beion are given by the equation 21001 E =-- 2 kJ/mole kJ/mole n=1, 2, 3.... - a. Calculate the energies in kJ/mole for the four lowest energy levels of the Be' ion. E = kJ/mole E = kJ/mole kJ/mole E = kJ/mole b. One of the most important transitions for the Beion involves a jump from the n = 2 to the n = 1 level. A for this transition equals E-E. where these two energies are obtained as in Part (a). Find the value of AE in kJ/mole. Find the wavelength in nm of the line emitted when this transition occurs: use Equation 4 to make the calculation AE = kJ/mole: 1 nm (continued on following page)

Answer 1

Given, E_n = - $\frac{21001}{n^{2}}$ KJ/mole

E₁ = - $\frac{21001}{1^{2}}$ KJ/mole. = - 21001 KJ/mole

E₂ = - $\frac{21001}{2^{2}}$ KJ/mole = - 5250.25 KJ/mole

E₃ = -   KJ/mole = - 2333.44 KJ/mole

21001 32

E₄ = - $\frac{21001}{4^{2}}$ KJ/mole = - 1312.56 KJ/mole.

b.

E₂ - E₁ = - 5250.25 - (-21001) KJ/mole

Or, $\Delta$ E = 15750.75 KJ/mole.

Now, energy per photon

E = $\frac{\Delta E}{N_A}$

N_A is avogadro's number = 6.02×10²³

Hence,

E = $\frac{15750750}{6.02\times10^{23}}$ = 2.62 × 10^-17

Now, E = hC/$\lambda$

Or 2.62×10^-17 = 6.626×10^-34 × 3×10⁸/$\lambda$

Or, $\lambda$ = (6.626×3×10^-26)/(2.62×10^-17)

Or, $\lambda$ = 7.58 ×10^-9 m

Or, $\lambda$ = 7.58 nm

Hence, wavelength = 7.58 nm.