Question

Lab Section: Experiment #11: The Atomic Spectrum of Hydrogen Advanced Study Assignment 1. Found in the gas phase, the beryllium trication, Be has an energy level formula analogous to that of the hydrogen atom, since both species have only one electron. The energy levels of the Belon are given by the equation E, = -2100-kl/mole n = 1.2.3. - Calculate the energies in kl/mole for the four lowest energy levels of the Belon. a kl/mole kl/mole kJ/mole kl/mole b. One of the most important transitions for the Belion involves a jump from the n = 2 to the na1 level. As for this transition equals E,-Ewhere these two energies are obtained in part (a). Find the value of Ein kl/mole. Find the wavelength in nm of the line emitted when this transition occurs, use equation 4 to make the calculations. AF = kJ/mole; 2 = 11-5

Answer 1

1.

a) Given that En = -21001/n2

Therefore,

E1 = -21001/12 = -21001kJ/mol

E2 = -21001/22 = -5250.25kJ/mol

E3 = -21001/32 = -2333.44kJ/mol

E4 = -21001/42 = -1312.56kJ/mol

b) Given $\Delta$ E = E2 - E1

           = -5250.25 - (-21001)

           = -5250.25 + 21001

           = 15750.75kJ/mol

   Also to find $\lambda$ we use the equation given in 4.

   $\lambda$ (in nm) = (1.19627 * 105)/$\Delta$E

           = (1.19627 * 105)/15750.75

           = 1.19627 / .1575075 = 7.595 nm.

c) (1) 4 --> 1

   (2) 3 --> 1

   (3) 4 --> 3

Calculations:

We can solve it by carefull observation and intuition. We got the value of $\lambda$ for the above problem as approx. 7.6 nm, where transition was from E2 to E1. We know that with increase in $\Delta$ E lower is the observed $\lambda$ . Thus to get a smaller value (as in case of (1) and (2) ) we tried to increase $\Delta$ E.

4 --> 1 : $\Delta$ E = E4 - E1 = 0.19688 * 105

$\lambda$(in nm) = (1.19627 * 105)/$\Delta$E

        = 1.19627 / 0.19688

        = 6.078 nm

3 --> 1 : $\Delta$ E = E3 - E1 = 0.1866756 * 105

$\lambda$(in nm) = (1.19627 * 105)/$\Delta$E

        = 1.19627 / 0.1866756

        = 6.411 nm

4 --> 3 : $\Delta$ E = E4 - E3 = 0.0102088 * 105

$\lambda$(in nm) = (1.19627 * 105)/$\Delta$E

        = 1.19627 / 0.0102088

        = 117.23 nm

PS : I would like to answer the second part, but its against the code. If there are more than 1 question only the first question should be answered.