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 at900. 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 \(2.56 \mathrm{L} \operatorname{In}\left(\mathrm{CH}_{3}\right)_{3}\) at 2.00 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 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 the moles of reactants using Ideal Gas Law

First, use the Ideal Gas Law (\(PV = nRT\)) to find the moles of each reactant. For In(CH3)3: \(n_{In(CH3)3} = \frac{PV}{RT} = \frac{(2.00\,\text{atm})(2.56\,\text{L})}{(0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}})(900\,\text{K})} = 0.00849\,\text{mol}\) For PH3: \(n_{PH3} = \frac{PV}{RT} = \frac{(3.00\,\text{atm})(1.38\,\text{L})}{(0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}})(900\,\text{K})} = 0.00861\,\text{mol}\)

02

Find the limiting reactant

Since the stoichiometry of the reaction shows that 1 mole of In(CH3)3 reacts with 1 mole of PH3 to produce 1 mole of InP: Limiting reactant = Min(\(n_{In(CH3)3}, n_{PH3}\)) = \(n_{In(CH3)3}\) = 0.00849 mol

03

Calculate the theoretical moles of InP and multiply by reaction yield

According to the stoichiometry, we produce 1 mole of InP per mole of limiting reactant. Multiply the theoretical moles of InP by the reaction yield (87 %): Moles of InP produced = \(0.00849\,\text{mol}\times (0.87) = 0.00739\,\text{mol}\)

04

Convert moles of InP to mass

Finally, we multiply the moles of InP by the molar mass of InP to find the mass: Mass of InP = \(0.00739\,\text{mol} \times (114.8\,\frac{\text{g}}{\text{mol}})\) = 0.849\, g Answer (a): The mass of InP(s) produced is 0.849 g. b. Wavelength of the light and visibility

05

Calculate the wavelength using energy

To find the wavelength (\(\lambda\)) of the light emitted, we can use the energy (\(E\)) and Planck's constant (\(h\)) and the speed of light (\(c\)) as follows: \(E = \frac{hc}{\lambda}\) So, \(\lambda = \frac{hc}{E}= \frac{(6.63\times10^{-34}\,\text{J}\cdot \text{s})(3.00\times 10^{8}\,\frac{\text{m}}{\text{s}})}{2.03\times10^{-19}\,\text{J}} = 9.79\times 10^{-7}\,\text{m}\)

06

Determine if the wavelength is visible to the human eye

The wavelength of the emitted light is \(9.79\times 10^{-7}\,\text{m}\) or \(979\,\text{nm}\). The range of visible light for the human eye is typically 380 - 750 nm. Since the wavelength is outside this range, the light emitted is not visible to the human eye. Answer (b): The wavelength of the light emitted is \(979\,\text{nm}\), and it is not visible to the human eye. c. Doping type

07

Identify the type of doping based on the electron configuration

The electron configuration of the replacing atom is [Kr] 5s^2 4d^10 5p^4. Since it has 4 valence electrons, which is 1 less than Phosphorus, it will create an extra hole by one less bond. Therefore, this type of doping is p-type doping. Answer (c): The doping is p-type doping.

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