Question

1. a) Calculate the ionisation energy of rubidium per atom, if light of wavelength 5.84  108 m produces electrons with a speed of 2.450  106 ms1. (5)
[Hint: Assume that the threshold frequency refers to the frequency corresponding to the ionisation energy.]
b) Assume that the electron in Li2+ ion is in third orbit. Calculate (5)
i) the radius of the orbit, and
ii) the total energy of the electron
[Hint: Li2+ ion also has atomic spectra similar to hydrogen atom. while applying relevant equations, use Z = 3.]

Answer 1

Answer on Question #62190, Chemistry / General Chemistry

a) Calculate the ionisation energy of rubidium per atom, if light of wavelength $5.84 \times 10^{-8} \mathrm{~m}$ produces electrons with a speed of $2.450 \times 10^{6} \mathrm{~ms}^{-1}$. [Hint: Assume that the threshold frequency refers to the frequency corresponding to the ionisation energy.

Solution:

Einstein's equation for the photoelectric effect

$$h\nu = K + W$$

Where, $h$ is the Planck constant and $\nu$ is the frequency of the incident photon. The $W$ is the work function. The kinetic energy $K$ of an ejected electron.

$$W = h\nu - K$$

Where,

$$\nu = \frac{c}{\lambda}$$ $$K = \frac{m\nu^2}{2}$$

Determine the energy:

$$W = h\frac{c}{\lambda} - \frac{m\nu^2}{2}$$ $$W = \frac{3 \cdot 10^{8} \mathrm{m/s} \times 6.626 \times 10^{-34} \mathrm{js}}{5.84 \cdot 10^{-8} \mathrm{m}} - \frac{9.1 \cdot 10^{-31} \mathrm{kg} \times (2.450 \cdot 10^{6} \mathrm{ms}^{-1})^2}{2} = 6.6 \cdot 10^{-19} \mathrm{J}$$

This is the energy required to ionize 1 atom of Rb

Answer: $6.6 \cdot 10^{-19} \mathrm{J}$

b) Assume that the electron in $\mathrm{Li}^{2+}$ ion is in third orbit. Calculate: i) the radius of the orbit, and ii) the total energy of the electron. [Hint: $\mathrm{Li}^{2+}$ ion also has atomic spectra similar to hydrogen atom. while applying relevant equations, use $Z = 3$.]

Solution:

Radius of atom

$$E_n = -E_0 \frac{Z^2}{n^2} = -13.6 \mathrm{eV} \frac{9}{n^2}$$

The first $\mathrm{Li}^{2+}$ level that have the same energy as hydrogen atom is: $n = 3$, $E_3 = -13.6 \mathrm{eV}$.

The radius of the third orbit would be

$$r = n^2 \frac{r_1}{Z} = 9 \frac{0.5 \cdot 10^{-10} \mathrm{m}}{3} = 1.5 \cdot 10^{-10} \mathrm{m} = 1.5 A^{\circ}$$

The electron's total energy

$$E = \frac{Z k e^2}{2 r} = \frac{3 \times 9 \cdot 10^9 \times (1.6 \cdot 10^{-19})^2}{2 \times 1.5 \cdot 10^{-10}} = 23.04 \cdot 10^{-19} \mathrm{J}$$

Answer: $1.5 A^{\circ}$; $23.04 \cdot 10^{-19} \mathrm{J}$

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