Indium(III) phosphide is a semiconducting material that has been frequently used in lasers, light-emitting diodes (LED), and fiber-optic devices. This material can be synthesized at 900\. \(\mathrm{K}\) according to the following reaction: $$ \operatorname{In}\left(\mathrm{CH}_{3}\right)_{3}(g)+\mathrm{PH}_{3}(g) \longrightarrow \operatorname{InP}(s)+3 \mathrm{CH}_{4}(g) $$ a. If \(2.56 \mathrm{~L} \mathrm{In}\left(\mathrm{CH}_{3}\right)_{3}\) at \(2.00 \mathrm{~atm}\) is allowed to react with \(1.38 \mathrm{~L} \mathrm{PH}_{3}\) at \(3.00 \mathrm{~atm}\), what mass of \(\operatorname{In} \mathrm{P}(s)\) will be produced assuming the reaction has an \(87 \%\) yield? b. When an electric current is passed through an optoelectronic device containing InP, the light emitted has an energy of \(2.03 \times 10^{-19} \mathrm{~J}\). What is the wavelength of this light and is it visible to the human eye? c. The semiconducting properties of \(\mathrm{InP}\) can be altered by doping. If a small number of phosphorus atoms are replaced by atoms with an electron configuration of \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\), is this n-type or \(\mathrm{p}\) -type doping?

Step by step solution

01

Calculate moles of reactants

Use the ideal gas law \(PV = nRT\) to calculate the moles of each reactant. For In(CH3)3, we have: \(P = 2.00 \: atm \), \(V = 2.56 \: L \), \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K} \), \(T = 900K\) \(n_{In(CH_{3})_{3}} = \frac{PV}{RT} = \frac{(2.00 \: atm)(2.56 \: L)}{(0.0821 \frac{L \cdot atm}{mol \cdot K})(900K)}\) Similarly for PH3: \(P = 3.00 \: atm \), \(V = 1.38 \: L\), \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\), \(T = 900K\) \(n_{PH_{3}} = \frac{PV}{RT} = \frac{(3.00 \: atm)(1.38 \: L)}{(0.0821 \frac{L \cdot atm}{mol \cdot K})(900K)}\)

02

Determine the limiting reactant

Calculate the mole ratio for the reaction and compare it to the given stoichiometry (1:1) to find the limiting reactant: \(\frac{n_{In(CH_{3})_{3}}}{n_{PH_{3}}} = \frac{0.0854 \: mol}{0.0456 \: mol}\) Since In(CH3)3 and PH3 react in a 1:1 ratio, PH3 is the limiting reactant.

03

Find theoretical moles and mass of InP

Use the stoichiometry to find the moles of InP produced theoretically: \(1\: mol\: PH_{3} \rightarrow 1\:mol\: InP\) \(n_{InP(theoretical)} = n_{PH_{3}} \) Now, find the mass using molar mass of InP (\(M_{InP} = 115 + 31 = 146 \: g/mol\)): \(mass_{InP(theoretical)} = n_{InP(theoretical)} M_{InP}\)

04

Calculate the actual mass of InP produced

Using the given reaction yield of 87%, calculate the actual mass of InP produced: \(mass_{InP(actual)} = mass_{InP(theoretical)} \times 0.87\) b. Wavelength of emitted light

05

Calculate the frequency of the light

Using the energy per photon \(E = 2.03 \times 10^{-19} J\), calculate the frequency (\(ν\)) using the Planck's constant (\(h = 6.63 \times 10^{-34} Js\)): \(E = hν\) \(ν = \frac{E}{h}\)

06

Calculate the wavelength of the light

Calculate the wavelength (\(λ\)) using the speed of light in a vacuum (\(c = 3.0 \times 10^8 m/s\)): \(c = λν\) \(λ = \frac{c}{ν}\) Determine if the wavelength lies within the visible light range (380 nm to 750 nm). c. Type of doping

07

Identify the doping element

Examine the electron configuration given (\([Kr] 5s^{2} 4d^{10} 5p^{4}\)) to identify the doping element, which turns out to be Te (tellurium).

08

Determine the type of doping

Te has one more valence electron than P, and thus donates an excess electron to the crystal lattice. Hence, it is n-type doping. #Summary-Output#: a. The mass of InP with 87% yield is approximately \(6.30 g\). b. The wavelength of the emitted light is approximately \(975 nm\), which is not visible to the human eye (infrared region). c. The given doping is n-type doping.

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