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3 years ago

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Ultraviolet photoelectron spectroscopy (UPS) is an important analytical tool used to characterize the composition of materials i

n academic, industrial and national laboratories. The technique relies on the photoelectric effect where-in electrons are emitted from a material when it absorbs ultraviolet photons. Make a plot of the kinetic energy of the photo-emitted electrons versus the frequency of the incident light as the frequency is varied across the transition where photon energy exceeds the ionization energy (IE) of a material being examined by UPS. Label you axes.

1 answer:

Setler [38]3 years ago

4 0

Answer:

The equation for the Kinetic Energy, $KE_e$ , of the photo-emitted electrons is given as follows;

$KE_e = h\cdot f- BE$

Please find attached the required plot of the kinetic energy versus the frequency

Explanation:

The equation for the Kinetic Energy, $KE_e$ , of the photo-emitted electrons is given as follows;

$KE_e = h\cdot f- BE$

Where:

h = Planck's constant = 6.63×10⁻³⁴ J·s

f = Frequency

BE = Binding Energy = 2.71 eV for Calcium metal

f₀ = Threshold frequency for the material

The ionization energy is the energy required to free an electron from an isolated atom

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Answer:

see explaination

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6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)

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Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

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Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

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$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

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C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

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Now put all the given values in the above formula, we get:

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

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Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

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7nadin3 [17]

Answer:

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upon reversing ( 2 ) equation

$N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$ Kc = 1/2.16e - 24

now adding 1 and reversed equation (2)

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we get ,

$N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)$ Kc = 1.54e-31 × 1/2.61e - 24

equilibrium constant of equation (3) is -

Kc = 1.54e - 31 / 2.61e - 24

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Hope this helped!

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