Question

Using Rydberg's equation for the energy levels (E = -R/n^2) and the wavelength and the c=(lambda)v, E=hv equations, SHOW/ PROVE how E(red band in H-spectrum, lambda = 656 nm) = (Delta or change/ difference) E (3 --> 2)?

Answer 1

Answer on Question #56439 - Chemistry - Other

Question:

Using Rydberg's equation for the energy levels $(E = -R/n^2)$ and the wavelength and the $c=(lambda)v$, $E=hv$ equations, SHOW/ PROVE how $E$(red band in H-spectrum, lambda = 656 nm) = (Delta or change/ difference) $E$ (3 --> 2)?

Answer:

Let's calculate the energy values for the energy levels 3 and 2:

$$E_n = - \frac{R}{n^2}, \quad E_3 = - \frac{2.178 \cdot 10^{-18} (J)}{3^2} = -2.42 \cdot 10^{-19} J$$ $$E_2 = - \frac{2.178 \cdot 10^{-18} (J)}{2^2} = -5.45 \cdot 10^{-19} J$$

Then, the difference between the energy of $3^d$ and $2^{nd}$ levels is:

$$\Delta E = E_3 - E_2 = (-2.42 + 5.45) \cdot 10^{-19} = 3.03 \cdot 10^{-19} J$$

The wavelength, corresponding to this energy change is:

$$E = h\nu = \frac{hc}{\lambda}, \lambda = \frac{hc}{E} = \frac{6.63 \cdot 10^{-34} (J \cdot s) \cdot 3 \cdot 10^8 (m \cdot s^{-1})}{3.03 \cdot 10^{-19} (J)} = 6.56 \cdot 10^{-7} m = 656 \text{ nm}$$

www.AssignmentExpert.com