Question

3. a) Using the values of n calculated in #1 and the Bohr equation, calculate the energy change (in J) for each line in the hydrogen spectrum. Complete the following table. Show your work in the space below the table. b) Use colored pencils to sketch the coresponding lines in the diagram on the right. level energy Wavelength (nm) Initial Final Energy Change (J) Color n 410 434 |Ч16 656 2. 2. -n= 2 2 3 2 Energy Level Diagram 4. How does the wavelength change as energy increases?

Answer 1

According to Bohr's theory, when an electron jumps from a higher orbital (n2) to lower orbital (n1), then it emits energy. Agian higher the orbital, higher is its energy. Thus if E2 and E1 are the higher and lower orbitals respectively, then the energy difference between them can be calculated from the formula-

ΔE = E2 - E1 = −R_H *(1/n₂² − 1/n₁²)

where R_H = Rydberg constant = 2.18×10⁻¹⁸J

Now putting the above formula-

Wave length (nm)	n1	n2	Color	ΔE (J)
410	6	2	Purple	−RH *(1/n22 − 1/n12) = −(2.18×10−18J) *(1/62 − 1/22) = −(2.18×10−18J) *(1/36− 1/4) = 0.484 * 10−18J = 4.84 * 10−19 J
434	5	2	Blue	−RH *(1/n22 − 1/n12) = −(2.18×10−18J) *(1/52 − 1/22) = −(2.18×10−18J) *(1/25− 1/4) = 0.4578‬ * 10−18J = 4.578 * 10−19 J
486	4	2	Green	−RH *(1/n22 − 1/n12) = −(2.18×10−18J) *(1/42 − 1/22) = −(2.18×10−18J) *(1/16− 1/4) = 0.4087 * 10−18J = 4.087 * 10−19 J
656	3	2	Red	−RH *(1/n22 − 1/n12) = −(2.18×10−18J) *(1/32 − 1/22) = −(2.18×10−18J) *(1/9− 1/4) = 0.3027 * 10−18J = 3.027 * 10−19 J

4- From the above calculation, we can clearly see that as the wave length increases, the energy difference decreases. That means orbitals with lower energy has higher wave lengths and vice versa.