Question

what is the ionization energy for an atom (in kilojoules per mole) if the ejected electron has a velocity of 4.72×10^5 m s^-1 when ionized with a 179.3 nm wavelength laser. note that the mass of an electron is 9.1094×10^-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 178.6 nm shines on a piece of aluminum metal? the work function for aluminum is 4.08 ev. the mass of an electron is 9.1094×10^-31 kg and 1 ev = 1.602×10^-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 for the first - part

Use the formula $E = h
u=\frac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\ J\cdot s$, $c = 3\times10^{8}\ m/s$ and $\lambda = 179.3\times10^{-9}\ m$.
$E_{photon}=\frac{6.626\times10^{-34}\times3\times10^{8}}{179.3\times10^{-9}}= 1.109\times10^{-18}\ J$

Step2: Calculate the kinetic energy of the ejected electron for the first - part

Use the formula $K.E.=\frac{1}{2}mv^{2}$, where $m = 9.1094\times10^{-31}\ kg$ and $v = 4.72\times10^{5}\ m/s$.
$K.E.=\frac{1}{2}\times9.1094\times10^{-31}\times(4.72\times10^{5})^{2}=1.019\times10^{-19}\ J$

Step3: Calculate the ionization energy for the first - part

By the photoelectric effect equation $E_{photon}=I.E.+K.E.$, so $I.E.=E_{photon}-K.E.$.
$I.E.=(1.109\times10^{-18}- 1.019\times10^{-19})\ J = 1.0071\times10^{-18}\ J$
For 1 mole of atoms, multiply by Avogadro's number $N_{A}=6.022\times10^{23}\ mol^{-1}$ and convert to kJ/mol.
$I.E.=\frac{1.0071\times10^{-18}\times6.022\times10^{23}}{1000}=606.5\ kJ/mol$

Step4: Calculate the energy of the incident photon for the second - part

Use $E = \frac{hc}{\lambda}$, with $\lambda = 178.6\times10^{-9}\ m$, $h = 6.626\times10^{-34}\ J\cdot s$ and $c = 3\times10^{8}\ m/s$.
$E_{photon}=\frac{6.626\times10^{-34}\times3\times10^{8}}{178.6\times10^{-9}}=1.103\times10^{-18}\ J$

Step5: Convert the work - function to joules for the second - part

Given $\Phi = 4.08\ eV$, convert to joules using $1\ eV=1.602\times10^{-19}\ J$.
$\Phi = 4.08\times1.602\times10^{-19}\ J = 6.536\times10^{-19}\ J$

Step6: Calculate the kinetic energy of the ejected electron for the second - part

By the photoelectric effect equation $K.E.=E_{photon}-\Phi$.
$K.E.=(1.103\times10^{-18}-6.536\times10^{-19})\ J = 4.494\times10^{-19}\ J$

Step7: Calculate the speed of the ejected electron for the second - part

Since $K.E.=\frac{1}{2}mv^{2}$, then $v=\sqrt{\frac{2K.E.}{m}}$.
$v=\sqrt{\frac{2\times4.494\times10^{-19}}{9.1094\times10^{-31}}}=9.90\times10^{5}\ m/s$

Answer:

Question 9: 606.5
Question 10: 9.90, 5