4.1 - Practice problem answers

Khoi Tran

$$E = h\nu = \frac{hc}{\lambda} = \frac{hc}{1.0 \cdot 10^{-9} \text{m}} = 2.0 \cdot 10^{-16} \text{J}$$

$$E_{\text{photon}} = hv = \frac{hc}{\lambda} = \phi + \frac{1}{2}mu^2$$

$$u = \sqrt{2 \frac{E_{\text{photon}^{-\phi}}}{m}} = \sqrt{2 \frac{\frac{hc}{\lambda} -\phi}{m}}$$

Note: It is also correct to calculate $\frac{hc}{\lambda}$ before performing algebraic transformations.

a) $$u = \sqrt{2 \frac{\frac{hc}{340 \cdot 10^{-9}} - 4.94 \cdot 10^{-19}}{9.11*10^{-31}}} = 8.6 \cdot 10^5 \text{ms}^{-1}$$

b) $$u = \sqrt{2 \frac{\frac{hc}{850 \cdot 10^{-9}} - 4.94 \cdot 10^{-19}}{9.11*10^{-31}}} = \text{SYNTAX ERROR}$$ At this stage, you will notice that the correct answer is “SYNTAX ERROR”. This is because $\phi > E_{\text{photon}}(=\frac{hc}{\lambda})$. Therefore, electrons will not be ejected from the surface when light with wavelength 850 nm is shone onto it.

a) We know that the atomic number of hydrogen is 1, so Z = 1. The spectral line of hydrogen with the second-highest wavelength of the Balmer series will have the second-lowest energy (because $E=\frac{hc}{\lambda}$, and energy is inversely proportional to wavelength). Therefore, this line corresponds to the drop from n = 4 to n = 2.

We can now solve for R: $$\frac{1}{493 \cdot 10^{-9} \text{m}} = R \cdot (1)^2 \cdot (\frac{1}{2^2} - \frac{1}{4^2})$$

$$R = 1.082 \cdot 10^7 \text{m}^{-1}$$

b) Because energy of an EM wave is inversely proportional to its wavelength, the shortest possible wavelength carries the highest possible energy. In other words, the electron will fall from $n_2 \to \infty$ to $n_1 = 1$.

$$\frac{1}{\lambda} = \lim_{n_2 \to \infty} R_H (\frac{1}{1^2} - \frac{1}{n_2^2} = R_H)$$

$$\lambda = \frac{1}{R_H} = \frac{1}{1.082 \cdot 10^7} = 9.244 \cdot 10^{-8} \text{m} = 92.42 \text{nm}$$