Question

Atkins Exercise 9.3(b): When ultraviolet radiation of wavelength 58.4 nm from a helium lamp is directed on to a sample of xenon, electrons are ejected with a speed of 1.79 Mm s 1. Calculate the ionization energy of xenon. Atkins Exercise 9.9(a) (Extra Credit): Calculate the average kinetic and potential energies of an electron in the ground state of a hydrogen atom. Atkins Problem 9.18 (Extra Credit: The “size” of an atom is sometimes considered to be measured by the radius of a sphere that contains 90% of the charge density of the electrons in the outermost occupied orbital. (a) Considering that the “90% criterion” is an arbitrary one, make a graph for hydrogen in its ground state how its effective "size” varies with the % electron density considered. (b) Use your graph in part (a) with the 90% criterion to determine an effective “size” of hydrogen in its ground state.

Answer 1

According to HOMEWORKLIB RULES, I am authorized to attempt at most one question. As you haven't specified the question to be answered, I will answer the first one.

When a incident photon strikes an electron with energy greater than electron's ionization energy, it transfers it's energy to the electron of which one part of energy is used to remove that electron from the atom (ionization energy) and other part is used to impart kinetic energy to it.

$\therefore Energy \hspace{0.2cm} of \hspace{0.2cm} radiation = I.E + K.E$

where I.E is ionization energy and K.E is Kinetic energy

$\therefore I.E = Energy \hspace{0.2cm} of \hspace{0.2cm} radiation - K.E$

Energy of radiation = h $\nu$

where h is Planck's constant = 6.626 * 10^-34 J - sec

and $\nu$ is the frequency of radiation which = $c / \lambda$(speed of radiation wave / wavelength of radiation)

As UV ray is a type of electromagnetic radiation

Therefore, speed of UV ray = speed of light = 3 * 10⁸ m/s

$\therefore \nu = c/ \lambda = \frac{3 \times 10^8}{58.4 \times 10^{-9}}$

$\therefore \nu = 5.13 \times 10^{15} /sec$

$\therefore Energy \hspace{0.2cm} of \hspace{0.2cm} radiation = h\nu =6.626 \times 10^{-34} \times 5.13 \times 10^{15}$

$\therefore Energy \hspace{0.2cm} of \hspace{0.2cm} radiation = 34 \times 10^{-19} \hspace{0.2cm}J = 3.4 \times 10^{-18} \hspace{0.2cm}J$

$\therefore K.E = \frac{1}{2}mv^2$

where m is the mass of electron = 9.1 * 10^-31 kg and v is it's velocity = 1.79 * 10⁶ m/s (GIVEN)

$\therefore K.E = \frac{1}{2}\times 9.1 \times 10^{-31}\times (1.79 \times 10^6)^2$

$\therefore K.E = 1.46 \times 10^{-18} \hspace{0.2cm}J$

$\therefore I.E = Energy \hspace{0.2cm} of \hspace{0.2cm} radiation - K.E$

Putting the values : -

$\therefore I.E = 3.40 \times 10^{-18} - 1.46 \times 10^{-18}$

$\Rightarrow I.E = 1.94 \times 10^{-18} \hspace{0.2cm}J$

This is the energy required to knockout one electron.

For 1 mole of electrons i.e. 6.022 * 10²³ electrons

$\Rightarrow I.E = 1.94 \times 10^{-18} \times 6.022 \times 10^{23} \hspace{0.2cm}$

$\Rightarrow I.E = 1168.27 \hspace{0.2cm} kJ/mol$

Therefore, ionization energy of xenon = 1168.27 kilo Joules / mole