Question

question 9
1 pts
what is the ionization energy for an atom (in kilojoules per mole) if the ejected electron has a velocity of 4.65 x 10^5 m s^-1 when ionized with a 213.7 nm wavelength laser. note that the mass of an electron is 9.1094 x 10^-31 kg. report your answer to 1 decimal place and do not include units in your answer.

question 10
1 pts
the photoelectric work function, φ, is the minimum energy needed to eject an electron from the metal by irradiating it with light. determine the speed (in meters per second) of the ejected electron when light of wavelength 295.8 nm shines on a piece of aluminum metal? the work function for aluminum is 4.08 ev. the mass of an electron is 9.1094 x 10^-31 kg and 1 ev = 1.602 x 10^-19 j. report your answer in proper scientific notation to 3 sig figs by filling in the following blanks with units shown last.

Explanation:

Step1: Calculate the energy of the incident photon

The energy of a photon is given by $E = h
u=\frac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\ J\cdot s$, $c = 3\times 10^{8}\ m/s$ and $\lambda$ is the wavelength of the light.
For $\lambda = 213.7\ nm=213.7\times 10^{-9}\ m$, $E_{photon}=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{213.7\times 10^{-9}}\ J$
$E_{photon}=\frac{19.878\times 10^{-26}}{213.7\times 10^{-9}}\ J\approx9.302\times 10^{-19}\ J$

Step2: Calculate the kinetic energy of the ejected electron

The kinetic energy of an electron is $K = \frac{1}{2}mv^{2}$, where $m = 9.1094\times 10^{-31}\ kg$ and $v = 4.65\times 10^{5}\ m/s$
$K=\frac{1}{2}\times9.1094\times 10^{-31}\times(4.65\times 10^{5})^{2}\ J$
$K = \frac{1}{2}\times9.1094\times 10^{-31}\times2.16225\times 10^{11}\ J$
$K\approx9.81\times 10^{-20}\ J$

Step3: Calculate the ionization energy per atom

By the photoelectric - effect equation $E_{photon}=E_{ionization}+K$
$E_{ionization}=E_{photon}-K$
$E_{ionization}=9.302\times 10^{-19}\ J - 9.81\times 10^{-20}\ J$
$E_{ionization}=(9.302 - 0.981)\times 10^{-19}\ J=8.321\times 10^{-19}\ J$

Step4: Convert ionization energy to kJ/mol

$1\ mol = 6.022\times 10^{23}$ atoms
$E_{ionization\ per\ mole}=E_{ionization}\times N_{A}$
$E_{ionization\ per\ mole}=8.321\times 10^{-19}\ J\times6.022\times 10^{23}$
$E_{ionization\ per\ mole}=499.19\times 10^{4}\ J/mol = 499.2\ kJ/mol$

for Question 10:

Step1: Convert the work - function from eV to J

Given $\Phi = 4.08\ eV$, and $1\ eV=1.602\times 10^{-19}\ J$
$\Phi=4.08\times1.602\times 10^{-19}\ J = 6.536\times 10^{-19}\ J$

Step2: Calculate the energy of the incident photon

For $\lambda = 295.8\ nm = 295.8\times 10^{-9}\ m$, $E_{photon}=\frac{hc}{\lambda}=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{295.8\times 10^{-9}}\ J$
$E_{photon}=\frac{19.878\times 10^{-26}}{295.8\times 10^{-9}}\ J\approx6.719\times 10^{-19}\ J$

Step3: Calculate the kinetic energy of the ejected electron

By the photoelectric - effect equation $E_{photon}=\Phi+K$
$K = E_{photon}-\Phi$
$K=6.719\times 10^{-19}\ J-6.536\times 10^{-19}\ J = 0.183\times 10^{-19}\ J=1.83\times 10^{-20}\ J$

Step4: Calculate the speed of the ejected electron

Since $K=\frac{1}{2}mv^{2}$, then $v=\sqrt{\frac{2K}{m}}$
$v=\sqrt{\frac{2\times1.83\times 10^{-20}\ J}{9.1094\times 10^{-31}\ kg}}$
$v=\sqrt{\frac{3.66\times 10^{-20}}{9.1094\times 10^{-31}}\ m^{2}/s^{2}}$
$v=\sqrt{0.402\times 10^{11}\ m^{2}/s^{2}}\approx6.34\times 10^{5}\ m/s$

Answer:

499.2