It takes 208.4 kJ of energy to remove one mole of electrons form the atoms on the surface of rubidium metal. If rubidium metal is irradiated with 254 nm light, what is the maximum velocity the released electrons can have?
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dravid1992sla dravid1992sla
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It takes 208.4 kJ of energy to remove 1 mole of electrons from the atoms on the surface of rubidium metal. If rubidium metal is irradiated with 254-nm light, what is the maximum kinetic energy the released electrons can have?
pantaposlovni94
pantaposlovni94
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John Darryl B. Lagdameo
John Darryl B. Lagdameo
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The maximum kinetic energy an electron can have is shown below:
\begin{aligned} \text{KE = hv – hv$_o$} \end{aligned}
KE = hv – hv
o
where KE is the kinetic energy, hv is the energy of an incident photon, and hv_o
o
is the energy needed to remove one mole of electrons.
2
It takes 208.4 kiloJoules of energy to remove one mole of electrons (hv_o
o
) from the atoms on the surface of rubidium metal. Since there are 6.022^{23}
23
electrons in one mole of, we can calculate the hv_o
o
a single electron, which is shown below:
\begin{aligned} \text{hv$_o$ per electron = } \dfrac {\text{ hv$_o$ per mole}}{\text{N$_A$}}\\ \text{hv$_o$ per electron = } \dfrac {\text{$\dfrac {\text{208.8 kJ}}{\text{mole}}$}}{\text{$\dfrac {\text{6.022x10$^{23}$}}{\text{mole}}$}}\\ \text{hv$_o$ per electron = 3.467286616x10$^{-22}$ kJ} \end{aligned}
hv
o
per electron =
N
A
hv
o
per mole
hv
o
per electron =
mole
6.022x10
23
mole
208.8 kJ
hv
o
per electron = 3.467286616x10
−22
kJ
The conversion of kilojoule to Joule is shown below:
\begin{aligned} \text{1 kiloJoule = 1000 Joules} \end{aligned}
1 kiloJoule = 1000 Joules
The hv_o
o
per electron in Joules is shown below:
\begin{aligned} \text{hv$_o$ per electron = 3.467286616x10$^{-22}$ kJ} \times \dfrac {\text{1000 J}}{\text{1 kJ}}\\ \text{ hv$_o$ per electron = 3.467286616x10$^{-19}$ J} \end{aligned}
hv
o
per electron = 3.467286616x10
−22
kJ×
1 kJ
1000 J
hv
o
per electron = 3.467286616x10
−19
J
3
The radium metal is irradiated with 254 nanometers of light. The conversion of nanometers to meters is shown below:
\begin{aligned} \text{1 nanometer = 1x10$^{-9}$ meter} \end{aligned}
1 nanometer = 1x10
−9
meter
The constant h is the Planck’s constant that is equal to 6.626x10^{-34}
−34
Joule-second, while c is the speed of light, which is 3x10^8
8
meter per second. The energy of an incident photon (hv) of the radium metal irradiated at 254 nanometers is shown below:
\begin{aligned} \text{E = } \dfrac {\text{hc}}{\text{$\lambda$}}\\ \text{E = } \dfrac {\text{(6.626x10$^{-34}$ Js)(3x10$^8$ $\dfrac {\text{m}}{\text{s}}$}}{\text{254 nm}} \times \dfrac {\text{1 nm}}{\text{1x10$^{-9}$ m}}\\ \text{E = 7.825984252x10$^{-19}$ J} \end{aligned}
E =
λ
hc
E =
254 nm
(6.626x10
−34
Js)(3x10
8
s
m
×
1x10
−9
m
1 nm
E = 7.825984252x10
−19
J
4
The maximum kinetic energy of the released electrons in the radium metal irradiated at 254 nm is calculated below:
\begin{aligned} \text{KE = hv – hv$_o$}\\ \text{KE = 7.825984252x10$^{-19}$ J - 3.467286616x10$^{-19}$ J}\\ \text{KE = 4.358697636x10$^{-19}$ J $\approx$ 4.36x10$^{-19}$ J} \end{aligned}
KE = hv – hv
o
KE = 7.825984252x10
−19
J - 3.467286616x10
−19
J
KE = 4.358697636x10
−19
J ≈ 4.36x10
−19
J
RESULT
4.36x10^{-19}
−19
J