Question

Indium(III) phosphide is a semiconducting material that has been frequently used in lasers, light-emitting diodes (LED), and fiberoptic devices. This material can be synthesized at 900. K according to the following reaction:

$In\left(CH3\right)3\left(g\right)+1PH3\left(g\right)\to 88nInP\left(s\right)+13CH4\left(g\right)$

a. If 2.56 L In(CH3)3 at 2.00 atm is allowed to react with 1.38 L PH3 at 3.00 atm, what mass of InP(s) will be produced assuming the reaction is 87% efficient?

b. When an electric current is passed through an optoelectronic device containing InP, the light emitted has an energy of 2.03 3 10219 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 [Kr]5s24d105p4, is this n-type or p-type doping?

Short answer

1. The mass of lnP is 7.1g.
2. $\lambda$=979nm, hence it is not visible to human eye.
3. N-type doping is used.

Step-by-step solution

Definition of semiconducting material

Between conductors, which are typically metals, and nonconductors, which are insulators, exist materials called semiconductors (such as most ceramics). Pure elements like silicon and germanium, as well as compounds like gallium arsenide and cadmium selenide, are the building blocks of semiconductors.

Finding the reaction of indium phosphide

(a)

The following reaction can be used to make indium (III) phosphide.

$\text{ln}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}{\text{(g) + PH}}_{\text{3}}\text{(g)}\to {\text{lnP(s) + 3CH}}_{\text{4}}\text{(g)}$

The ideal gas equation can be used to compute the number of moles in this process.

$\text{PV = nRT}$

P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature in the equation above.

Computing the number of moles

The ideal gas equation may be used to compute the amount of moles ${\mathrm{PH}}_3$of

and $\left(\text{In}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\right)$as follows:

From the calculated moles, we can see that the ${\mathrm{PH}}_3$ is the limiting reactant. Hence, the mass of InP formed will depend on the mass of ${\mathrm{PH}}_3$.

Obtaining the mass and molar mass

From the balanced chemical equation, we can see that 1 mol of ${\mathrm{PH}}_3$is equal to 1 mol of InP.

$$\begin{gathered} n\left(\text{lnP}\right)\text{=n}\left({\text{PH}}_3\right) \\ \text{=0.056mol} \end{gathered}$$

The mass of $InP$can be calculated from the number of moles of InP and molar mass of InP.

But, since the efficiency of the reaction is 87%, the final mass of InP is:

$\text{mass}\left(\text{lnP}\right)\text{=8.164g×0.87=7.1g}$

Therefore, the mass of lnP is 7.1g.

Finding the wavelength of light

(b) When an electric current is sent through an optoelectronic device containing InP, we need to determine the wavelength of the light emitted.

We know that the amount of energy emitted is $2.30\times {10}^{-19}J$

The relation between energy and wavelength is as follows:

$\text{E=}\frac{\text{hc}}{\lambda }$

In the equation shown above, h is Planck's constant; c is the speed of light and$\lambda$ is the wavelength of light.

Now, Calculating wavelength of light

The energy of light is 2.30*10­­­^-19J.

The calculation of Wavelength of light is as follows:

$$\begin{gathered} \text{λ=}\frac{\text{hc}}{E} \\ \text{λ=}\frac{\left({\text{6.626×10}}^{\text{-34}}\text{Js}\right)\left({\text{3×10}}^8\text{m/s}\right)}{{\text{2.03×10}}^{\text{-19}}\text{J}} \\ \text{λ=979nm} \end{gathered}$$

This wavelength corresponds to IR region.

Wavelengths between 380 to 750 nanometers are visible to the human eye. The human eye cannot see the infrared area.

Therefore, it is not visible to the human eye and =979nm.

Obtaining the doping type

c)

Doping is the process of introducing impurities into a semiconductor crystal in order to change its conductivity in a specific way.

A p-type semiconductor is formed when an intrinsic semiconductor is doped by a trivalent impurity. Thep indicates that the semiconductor contains a lot of holes or positively charged ions.

We get pentavalent impurities when we dope intrinsic material with them.

" n " stands for negative in n-type semiconductors. Negatively charged ions, or surplus electrons, are found in N-type semiconductors.

The electronic configuration of the element used for doping is: $\left[Kr\right]5s24d105p4$

The element belongs to Group 6 due to its electrical arrangement, which has one electron more than phosphorus.

The system will have free electrons if the new element is employed to replace phosphorus. As a result, the n-type semiconductor will be generated.

Therefore N-type doping is used.