4.2 - Practice problem answers

  1. x $$\Delta x \Delta p = m \Delta x \Delta \nu = \frac{h}{4\pi}$$ $$\Delta x = \frac{h}{4m\Delta v\pi} = \frac{6.626 \cdot 10^{-34} J s}{4 \cdot (9.11 \cdot 10^{-31} \text{kg})(\frac{299792458 m/s}{10}) \pi} = 1.9 \cdot 10^{-12} m = 1.9 pm$$

  2. Answers

a) The leftmost line has the shortest wavelength, therefore, it has the highest energy and must be associated with the relaxation of electron from n = ∞ to n = 1 (the ground state). Therefore,

$$E = \frac{hc}{\lambda} = \lim{n \to \infty} Z^2 R_H (\frac{1}{1^2} - \frac{1}{n^2}) = Z^2 R_H$$

Because Z = 2, the one-electron species must be $\ce{He+}$.

b) Because the rightmost line is associated with n = 2, the fourth line from the right is from the relaxation of the electron from n = 5 to n = 1.

$$\frac{hc}{\lambda} = 2^2 R_H (\frac{1}{1^2} - \frac{1}{5^2})$$

$$\lambda = \frac{hc}{4R_H (1 - \frac{1}{25})} = 23.7 nm$$

  1. $$\lambda = \frac{h}{mu} = \frac{6.626 \cdot 10^{-34} J s}{(0.412 kg)(83.2 \frac{km}{h} \cdot \frac{1h}{3600s} \cdot \frac{1000 m}{1 km})} = 6.96 \cdot 10^{-35} m$$

  2. It seems like there is barely enough information to solve this problem. However, we can use equations we learned in the past to our advantage. Firstly, we are trying to calculate the wavelength of an electron. This means we will use De Broglie’s equation. However, we need to know the speed of the electron. We are given the energy of the electron, but we also know the mass of the electron thanks to the equation sheet. Therefore, we can use the equation that describes the kinetic energy of an object to find speed, and insert it into De Broglie’s equation (sheet also provides the value for Planck’s constant).

$$K_E = \frac{1}{2} mu^2$$ $$u = \sqrt{\frac{K_E}{\frac{1}{2}m}}$$ $$u = \sqrt{\frac{1.3616 \cdot 10^{-16}}{\frac{1}{2} (9.1093837 \cdot 10^{-31})}}$$ $$u = 17290011.07$$

$$\lambda = \frac{h}{mu}$$ $$\lambda = \frac{6.62607015 \cdot 10^{-34}}{(9.109387 \cdot 10^{-34})(17290011.07)}$$ $$\lambda = 4.2069 \cdot 10^{-11} m/s$$ or $$\lambda = 42.069 pm$$

Therefore, the wavelength of the electron is $4.2069 x 10^{-11}$ m/s, or $42.069$ pm.

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