Boost Your Understanding of De Broglie Equation with Our Examples

The quantity which is same for both the electron and the proton.

A proton with a speed of 0.9 m/s is directed through a double slit with slit separation 0.4mm. An array of detectors is placed 8m away from the slits. How far off axis is the 2nd minimum in the intensity pattern?
Group of answer choices
$4.4mm$
$4.4cm$
$13.2cm$
$1.32cm$
none of the above

How do you calculate the velocity of an electron if the debroy wavelength is given.

To determine
Which one have shorter wavelength when both proton and electron travels with same speed.

In de Broglie’s equation, he shows the relationship between the wavelength of a light wave and its momentum. Which of the following is true about this relationship?
wavelength is inversely proportional to the square of momentum
wavelength is directly proportional to momentum
wavelength is directly proportional to the square of momentum
wavelength is inversely proportional to momentum

de Broglie wavelength (equation) is is the wavelength, $\mathrm{\lambda <!--\; \lambda \; -->}$, associated with a massive particle and is related to its energy.
A. True
B. False

Find the de Broglie wavelength for a baseball weighing 149g that is moving at 101mph (fastball).

John Isner holds the ATP's (Association of Tennis Professionals) official record for the fastest serve at 253 km/h. Calculate the de Broglie wavelength of a tennis ball with the standard weight of 58.8 grams. Give your answer in ${10}^{-<!--\; -\; -->34}m$ (for example, if your answer is $5.6\times <!--\; \times \; -->{10}^{-<!--\; -\; -->34}$ m, then enter 5.6). This should give you a good idea of why one cannot expect to observe interference affects using tennis balls rather then, for example, electrons.

The speed of electron.

At what velocity does a proton have a 6.00-fm wavelength (about the size of a nucleus)? Assume the proton is nonrelativistic. (1 femtometer $={10}^{-15}m.$ )

JC-23) de Broglie Waves A photon and a particle have the same wavelength.
a) How do their linear momenta compare?
b) How does the photon's energy compare to the particle's total energy?
c) How does the photon's energy compare to the particle's kinetic energy?

Why is the de Broglie equation as well as the Schrodinger equation is correct for massive particle?
Starting from special relativity, here I see the de Broglie approximation is valid only if $m_0=0$
Derivation:
$E^2=P^2{C}^2+m_0^2{C}^4$. Here we put Plank-Einstein relation $E=h\mathrm{\nu <!--\; \nu \; -->}=h\frac{C}{\mathrm{\lambda <!--\; \lambda \; -->}}$. Finally,
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{\sqrt{P^2+m_0^2{C}^2}}(1)$
If $m_0=0$ then $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$ (de Broglie approximation).
Furthermore we know the Schrodinger equation was derived by assuming that the de Broglie approximation is true for all particles, even if $m_0\ne <!--\; \ne \; -->0$. But if we take special relativity very strictly then this approximation looks incorrect.
In addition, if we try to derive the Schrodinger equation from the exact relation found in '1', we find completely different equation. For checking it out, lets take a wave function-
$\mathrm{\Psi <!--\; \Psi \; -->}=A{e}^{i(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}}x-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}t)}=A{e}^{i(\frac{\sqrt{P^2+m_0^2{C}^2}}{\mathrm{\hslash <!--\; \hslash \; -->}}x-<!--\; -\; -->\frac{E}{\mathrm{\hslash <!--\; \hslash \; -->}}t)}$ (putting $\mathrm{\lambda <!--\; \lambda \; -->}$ from '1', $\frac{h}{2\mathrm{\pi <!--\; \pi \; -->}}=\mathrm{\hslash <!--\; \hslash \; -->}$ and $E=\mathrm{\hslash <!--\; \hslash \; -->}\mathrm{\omega <!--\; \omega \; -->}$).
Then, $\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}=-<!--\; -\; -->\frac{P^2+m_0^2{C}^2}{{\mathrm{\hslash <!--\; \hslash \; -->}}^2}\mathrm{\Psi <!--\; \Psi \; -->}=-<!--\; -\; -->\frac{E^2}{C^2{\mathrm{\hslash <!--\; \hslash \; -->}}^2}\mathrm{\Psi <!--\; \Psi \; -->}$
$⟹<!--\; ⟹\; -->E^2\mathrm{\Psi <!--\; \Psi \; -->}=-<!--\; -\; -->C^2{\mathrm{\hslash <!--\; \hslash \; -->}}^2\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}(2)$
Again, $\frac{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}=-<!--\; -\; -->i\frac{E}{\mathrm{\hslash <!--\; \hslash \; -->}}\mathrm{\Psi <!--\; \Psi \; -->}⟹<!--\; ⟹\; -->E\mathrm{\Psi <!--\; \Psi \; -->}=-<!--\; -\; -->i\mathrm{\hslash <!--\; \hslash \; -->}\frac{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}$
Here we see operator $E=-<!--\; -\; -->i\mathrm{\hslash <!--\; \hslash \; -->}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}⟹<!--\; ⟹\; -->E^2=-<!--\; -\; -->{\mathrm{\hslash <!--\; \hslash \; -->}}^2\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2}{\mathrm{\partial <!--\; \partial \; -->}{t}^2}$
$⟹<!--\; ⟹\; -->E^2\mathrm{\Psi <!--\; \Psi \; -->}=-<!--\; -\; -->{\mathrm{\hslash <!--\; \hslash \; -->}}^2\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{t}^2}(3)$
Combining (2) and (3) we find the differential equation:
$\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}=\frac{1}{C^2}\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{t}^2}$
It is the Maxwell's equation, not the well known Schrodiner equation!
Therefore for the Schrodinger equation to exist, the de Broglie approximation must hold for $m_0\ne <!--\; \ne \; -->0.$ I see a clear contradiction here. Then why is the Schrodinger equation correct after all?

Why doesn't calculating the de Broglie wavelength work with h in eV·s?
I'm trying to calculate the de Broglie wavelength of a particle with a known momentum $p=980.93\mathrm{G}\mathrm{e}\mathrm{V}/\mathrm{s}$. The de Broglie relation is:
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
but I notice that if I use the value of $4.136\times <!--\; \times \; -->{10}^{-<!--\; -\; -->15}\mathrm{e}\mathrm{V}\cdot <!--\; \cdot \; -->\mathrm{s}$ for the Planck's constant, the answer that I get from the equation is in units of ${\mathrm{s}}^2$, not m. Specifically:
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{4.136\times <!--\; \times \; -->{10}^{-<!--\; -\; -->15}\mathrm{e}\mathrm{V}\cdot <!--\; \cdot \; -->\mathrm{s}}{980.93\mathrm{G}\mathrm{e}\mathrm{V}/\mathrm{s}}=4.25\times <!--\; \times \; -->{10}^{-<!--\; -\; -->27}{\mathrm{s}}^2$
which is not at all equivalent to the value that one gets when using $h=6.626\times <!--\; \times \; -->{10}^{-<!--\; -\; -->34}\mathrm{J}\cdot <!--\; \cdot \; -->\mathrm{s}$ instead (since $\mathrm{J}\cdot <!--\; \cdot \; -->\mathrm{s}$ can be converted into $\frac{\mathrm{k}\mathrm{g}\cdot <!--\; \cdot \; -->{\mathrm{m}}^2}{{\mathrm{s}}^2}\cdot <!--\; \cdot \; -->\mathrm{s}$).
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{6.626\times <!--\; \times \; -->{10}^{-<!--\; -\; -->34}\frac{\mathrm{k}\mathrm{g}\cdot <!--\; \cdot \; -->{\mathrm{m}}^2}{{\mathrm{s}}^2}\cdot <!--\; \cdot \; -->\mathrm{s}}{5.237\times <!--\; \times \; -->{10}^{-<!--\; -\; -->16}\frac{\mathrm{k}\mathrm{g}\cdot <!--\; \cdot \; -->\mathrm{m}}{\mathrm{s}}}=1.27\times <!--\; \times \; -->{10}^{-<!--\; -\; -->18}\mathrm{m}$
What's going on here?

Why is a relativistic calculation needed unless $pc$ is much smaller than the rest energy of a particle?
After introducing the de Broglie wavelength equation, my textbook gives a rather simple example where it asks to find the kinetic energy of a proton whose de Broglie wavelength is 1 fm. In the solution to this problem, it states that "A relativistic calculation is needed unless $pc$ for the proton is much smaller than the proton rest energy."
Could someone please explain why this is the necessary condition? I'm not sure what the quantity '$pc$' represents or means here. I know that for massless particles like photons, the total energy $E$ is equal to $pc$. I'm not sure what it means for particles having rest mass like protons.

Relativistic change of mass and De-broglie's equation
From De Broglie's equation
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{mv}$
We can see that $\mathrm{\lambda <!--\; \lambda \; -->}$ depends inversely on $m$
My question is:
If the $m$ in the equation indicates the kinetic mass of the body then from this relation
$m=\frac{m_0}{\surd (1-<!--\; -\; -->v^2/c^2)}$
$m$ would increase with $v$. Thus the $\mathrm{\lambda <!--\; \lambda \; -->}$ value should change doubly...first for the rise of $v$ and second for the rise of $m$
Or is the $m$ in De Broglie's equation representing rest mass, meaning that I am altogether wrong.
I am a layman in Physics and I stumbled on this question while understanding electron microscopy

Calculating an energy of an electron with known De Broglie wavelength (why can't we calculate it similar than we do it for a photon)
Lets say we have an electron with known De Broglie wavelength $\mathrm{\lambda <!--\; \lambda \; -->}$. Can anyone justify or explain why we calculate its energy $E$ using 1st the De Broglie relation $\mathrm{\lambda <!--\; \lambda \; -->}=h/p$ to get momentum $p$ and 2nd using the invariant interval to calculate $E$
$\begin{array}{rl}{p}^2{c}^2& =E^2-<!--\; -\; -->{E_0}^2\\ E& =\sqrt{p^2{c}^2+{E_0}^2}\end{array}$
Why we are not alowed to do it like we do it for a photon:
$\begin{array}{r}E=h\mathrm{\nu <!--\; \nu \; -->}=h\frac{c}{\mathrm{\lambda <!--\; \lambda \; -->}}\end{array}$
These equations return different results.

Where am I doing mistake of concept in deriving the De Broglie equation?
When I am deriving de Broglie wavelength for a relativistic particle using $E^2=m^2{c}^4+p^2{c}^2$ and equating with $E=\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}$, and then putting $p=kmv$, $k$ being relativistic factor, I am getting $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{kmc}$ instead of $\frac{h}{kmv}$.
Is there any mistake that i am doing with equating those 2 energy equations ? Or something else ?

Combination of de Broglie wavelength and mass–energy equivalence gone wrong?
I tried to combine the mass–energy equivalence for a particle with mass,
$E=\sqrt{(m{c}^2{)}^2+(pc{)}^2}=\sqrt{(m{c}^2{)}^2+(\mathrm{\gamma <!--\; \gamma \; -->}mvc{)}^2}$
with de Broglie wavelength,
$\mathrm{\lambda <!--\; \lambda \; -->}={\displaystyle \frac{h}{p}}={\displaystyle \frac{h}{\mathrm{\gamma <!--\; \gamma \; -->}mv}}.$
I get this equation:
$E={\displaystyle \frac{h{c}^2}{\mathrm{\lambda <!--\; \lambda \; -->}v}}.$
This does not seem right, since the equations suggest the energy increases as the speed slow down which is not the case. But I can't see what I did wrong, either. Can someone help me?