Real-Life Examples of Quantum Mechanic

De Broglie's Matter wave equation dividing by zero
I was just thinking about De Broglie's matter wave equation: $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$ where $p$ is the momentum of the object. But what if the object is at rest? Won't we be dividing by zero? What if we take the limit as momentum tends to zero, won't we start to get noticeable waves? Can someone please explain to me where I went wrong?

Deriving Planck's Constant from Wien's Displacement Law
So I'm reading an introductory book on Quantum Theory (David Park, 3rd ed.) and I am having trouble with the following question:
"According to Wien's displacement law, the wavelength ${\mathrm{\lambda <!--\; \lambda \; -->}}_m$ at which blackbody radiation at temperature T has its maximum intensity is given roughly by ${\mathrm{\lambda <!--\; \lambda \; -->}}_{m}T\simeq <!--\; \simeq \; -->$ 3 mm K. Assuming that the quantum energy at this temperature is of the order of kT where k is Boltzmann's constant, estimate the value of Planck's constant."
My attempt at a solution is as follows:
${\mathrm{\lambda <!--\; \lambda \; -->}}_{m}T\simeq <!--\; \simeq \; -->$ 3 mm K $,$$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{c}{\mathrm{\nu <!--\; \nu \; -->}}$$\to <!--\; \to \; -->$$\frac{cT}{\mathrm{\nu <!--\; \nu \; -->}}=3\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->3}mK$$\cdot <!--\; \cdot \; -->(\frac{k}{c})\to <!--\; \to \; -->$$\frac{kT}{\mathrm{\nu <!--\; \nu \; -->}}=1.38\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->34}Js$
$(\frac{1}{2}m{v}^2{)}_{max}=k{T}_{max}=h\mathrm{\nu <!--\; \nu \; -->}-<!--\; -\; -->e\mathrm{\varphi <!--\; \varphi \; -->}$
$\frac{k{T}_{max}}{\mathrm{\nu <!--\; \nu \; -->}}+\frac{e\mathrm{\varphi <!--\; \varphi \; -->}}{\mathrm{\nu <!--\; \nu \; -->}}=h$
$1.38\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->34}+\frac{e\mathrm{\varphi <!--\; \varphi \; -->}}{\mathrm{\nu <!--\; \nu \; -->}}=h$
At this point I get stuck, the orders of magnitude and units seem to be right but I'm not sure what to invoke/what is given in the question that can solve/eliminate this last term to get h, or if I'm even on the right track.

How do quasars produce light?
Wikipedia states that the radiation of quasars is partially nonthermal (i.e. not blackbody).
Well, what percentage of the total radiation isn't black body?
For some reason Wikipedia doesn't give a number for the temperature of the gas of the accretion disc of a quasar. I had to search for a while and tried many keywords on Google to get an answer of 80 million Kelvin. If someone knows better and/or can give me a range instead of a single number that would be nice.
Based on that number and an online calculator of wien's displacement law I got that quasars emit blackbody radiation mainly on the x-ray and gamma ray section of the spectrum. Is that true?

Inferring Properties of Stars from Masses and Radii
I have two questions related to inferring properties of stars from their masses and radii.
1.What properties of a star's spectrum could we deduce? In particular, do all stars emit like black bodies? If so, I'm guessing that one can calculate the star's effective temperature and use Wien's displacement law to find the peak of the black body spectrum.
2.How would the mass and radius influence Type 1 X-ray bursts?

Physics of an infrared thermometer
The thing about infrared thermometers that bugs me is how can you get the same temperature reading regardless of the distance to the object. Shouldn't there be a difference when measuring from two different standing points since energy flux density decreases with $\frac{1}{distanc{e}^2}$ and infrared thermometers work by focusing IR light on a thermopile, which then results in decreased (when measuring from further away) absorbed energy and therefore lower temperature and finally lower voltage across thermopile. Is there something I am getting wrong about this, or do IR thermometers make use of some other physics law like Wien's displacement law, by somehow measuring ${\mathrm{\lambda <!--\; \lambda \; -->}}_{peak}$ to determine the temperature?

The temperature of the surface of the sun
If I calculate the temperature of the Sun by replacing the effective frequency of the Sun (598 THz) in Planck's curve for Blackbody radiation, the result will be 10170 K. Still, in Wien's displacement law, the result will be 5778 K, is there an answer for this problem?

Showing Wien's Displacement Law from Wien's Law
Does anyone know how I would show that $\mathrm{\lambda <!--\; \lambda \; -->}\ast <!--\; \ast \; -->T$ is constant, using only Wien's Law? That $\mathrm{\rho <!--\; \rho \; -->}(\mathrm{\lambda <!--\; \lambda \; -->},T)=1/{\mathrm{\lambda <!--\; \lambda \; -->}}^5\ast <!--\; \ast \; -->f(\mathrm{\lambda <!--\; \lambda \; -->}T)$ I differentiated, but all I could get was $\mathrm{\lambda <!--\; \lambda \; -->}T=5f(\mathrm{\lambda <!--\; \lambda \; -->}T)/f^{\prime }(\mathrm{\lambda <!--\; \lambda \; -->}T)$, which I don't think means it's necessarily a constant.

Is Planck temperature really the highest temperature?
Actually I was learning about Wien's displacement law. It states that
$\mathrm{\lambda <!--\; \lambda \; -->}T=2.898\times {10}^{-<!--\; -\; -->3}mK$
This is actually a part of Planck's law where the Planck's constant originated.
Now Planck's temperature is given as
$T_p=\sqrt{\frac{h{c}^5}{2\mathrm{\pi <!--\; \pi \; -->}G{k}_b^2}}=1.416\times {10}^{32}K$
Now Planck's length is $1.616\times {10}^{-<!--\; -\; -->35}m$
Now since the smallest possible wavelength is Planck's length, we can say wavelength of the electromagnetic radiation is Planck's length (Assume the energy doesn't create a black hole).
Now according to Wien's displacement law,
$l_{p}T=2.898\times {10}^{-<!--\; -\; -->3}mK$
Now solving this we get $1.79\times {10}^{32}K$, which is higher than the actual Planck's temperature.
Since this displacement law is completely derived from Planck's law, it bit frustrated me. I'm a bit confused. Is it the limit of the displacement law or my flaw?
Please rectify this.
(Sorry if I made any mistake. I'm new to this one. Please explain my mistake. I'm glad to hear that.)