Find the de Broglie wavelength of a football of mass 0.4 kg traveling 23 m/s?

Jadon Camacho
2022-11-08
Answered

Find the de Broglie wavelength of a football of mass 0.4 kg traveling 23 m/s?

You can still ask an expert for help

Gwendolyn Alexander

Answered 2022-11-09
Author has **16** answers

The equation for the de Broglie wavelength is given by,

$\lambda =\frac{h}{mv}$

Here, h is the Planck's constant, m is the mass of the object and v is the speed of the object.

Substituting the values in the equation.

$\lambda =\frac{6.626\times {10}^{-34}J\cdot s}{(0.4kg)(23m/s)}\phantom{\rule{0ex}{0ex}}=7.20\times {10}^{-35}m$

$\lambda =\frac{h}{mv}$

Here, h is the Planck's constant, m is the mass of the object and v is the speed of the object.

Substituting the values in the equation.

$\lambda =\frac{6.626\times {10}^{-34}J\cdot s}{(0.4kg)(23m/s)}\phantom{\rule{0ex}{0ex}}=7.20\times {10}^{-35}m$

asked 2022-05-15

De Broglie wavelength of an electron

If an electron which already possesses some kinetic energy of $X\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$ is further accelerated through a potential difference of $X\prime \phantom{\rule{thinmathspace}{0ex}}\mathrm{V}$, then is it correct to state that the electron now has a total kinetic energy of $(X+X\prime )\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?

If an electron which already possesses some kinetic energy of $X\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$ is further accelerated through a potential difference of $X\prime \phantom{\rule{thinmathspace}{0ex}}\mathrm{V}$, then is it correct to state that the electron now has a total kinetic energy of $(X+X\prime )\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?

asked 2022-05-07

Direction of momentum given by the de Broglie relation

$p=mv$

where $m$ is the mass of an electron, and $v$ is its velocity. In this case, since $v$ is a vector, it's clear that the momentum will be also a vector.

However if the momentum is a vector quantity (and it is), what is the direction of the electron's momentum given by the de Broglie relation

$p=h/\lambda \phantom{\rule{0ex}{0ex}}p=\hslash k$

if the Planck constant $h$ is scalar and the wavelength $\lambda $ is also scalar. Similarly the reduced Planck constant $\hslash $ is scalar and the wavenumber $k=2\pi /\lambda $ is also scalar.

$p=mv$

where $m$ is the mass of an electron, and $v$ is its velocity. In this case, since $v$ is a vector, it's clear that the momentum will be also a vector.

However if the momentum is a vector quantity (and it is), what is the direction of the electron's momentum given by the de Broglie relation

$p=h/\lambda \phantom{\rule{0ex}{0ex}}p=\hslash k$

if the Planck constant $h$ is scalar and the wavelength $\lambda $ is also scalar. Similarly the reduced Planck constant $\hslash $ is scalar and the wavenumber $k=2\pi /\lambda $ is also scalar.

asked 2022-10-14

To determine

To Calculate: Momentum of photon by using equation $p=mv$

To Calculate: Momentum of photon by using equation $p=mv$

asked 2022-08-13

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

Group of answer choices

$4.4mm$

$4.4cm$

$13.2cm$

$1.32cm$

none of the above

asked 2022-11-03

10. 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?

1. wavelength is directly proportional to the square of momentum

2. wavelength is inversely proportional to momentum

3. wavelength is directly proportional to momentum

4. wavelength is inversely proportional to the square of momentum

1. wavelength is directly proportional to the square of momentum

2. wavelength is inversely proportional to momentum

3. wavelength is directly proportional to momentum

4. wavelength is inversely proportional to the square of momentum

asked 2022-10-02

Factor 2 in Heisenberg Uncertainty Principle: Which formula is correct?

Some websites and textbooks refer to

$\mathrm{\Delta}x\mathrm{\Delta}p\ge \frac{\hslash}{2}$

as the correct formula for the uncertainty principle whereas other sources use the formula

$\mathrm{\Delta}x\mathrm{\Delta}p\ge \hslash .$

Question: Which one is correct and why?

The latter is used in the textbook "Physics II for Dummies" (German edition) for several examples and the author also derives that formula so I assume that this is not a typing error.

This is the mentioned derivation:

$\mathrm{sin}\theta =\frac{\lambda}{\mathrm{\Delta}y}$

assuming $\theta $ is small:

$\mathrm{tan}\theta =\frac{\lambda}{\mathrm{\Delta}y}$

de Broglie equation:

$\lambda =\frac{h}{{p}_{x}}$

$\Rightarrow \mathrm{tan}\theta \approx \frac{h}{{p}_{x}\cdot \mathrm{\Delta}y}$

but also:

$\mathrm{tan}\theta =\frac{\mathrm{\Delta}{p}_{y}}{{p}_{x}}$

equalize $\mathrm{tan}\theta $:

$\frac{h}{{p}_{x}\cdot \mathrm{\Delta}y}\approx \frac{\mathrm{\Delta}{p}_{y}}{{p}_{x}}$

$\Rightarrow \frac{h}{\mathrm{\Delta}y}\approx \mathrm{\Delta}{p}_{y}\Rightarrow \mathrm{\Delta}{p}_{y}\mathrm{\Delta}y\approx h$

$\Rightarrow \mathrm{\Delta}{p}_{y}\mathrm{\Delta}y\ge \frac{h}{2\pi}$

$\Rightarrow \mathrm{\Delta}p\mathrm{\Delta}x\ge \frac{h}{2\pi}$

Some websites and textbooks refer to

$\mathrm{\Delta}x\mathrm{\Delta}p\ge \frac{\hslash}{2}$

as the correct formula for the uncertainty principle whereas other sources use the formula

$\mathrm{\Delta}x\mathrm{\Delta}p\ge \hslash .$

Question: Which one is correct and why?

The latter is used in the textbook "Physics II for Dummies" (German edition) for several examples and the author also derives that formula so I assume that this is not a typing error.

This is the mentioned derivation:

$\mathrm{sin}\theta =\frac{\lambda}{\mathrm{\Delta}y}$

assuming $\theta $ is small:

$\mathrm{tan}\theta =\frac{\lambda}{\mathrm{\Delta}y}$

de Broglie equation:

$\lambda =\frac{h}{{p}_{x}}$

$\Rightarrow \mathrm{tan}\theta \approx \frac{h}{{p}_{x}\cdot \mathrm{\Delta}y}$

but also:

$\mathrm{tan}\theta =\frac{\mathrm{\Delta}{p}_{y}}{{p}_{x}}$

equalize $\mathrm{tan}\theta $:

$\frac{h}{{p}_{x}\cdot \mathrm{\Delta}y}\approx \frac{\mathrm{\Delta}{p}_{y}}{{p}_{x}}$

$\Rightarrow \frac{h}{\mathrm{\Delta}y}\approx \mathrm{\Delta}{p}_{y}\Rightarrow \mathrm{\Delta}{p}_{y}\mathrm{\Delta}y\approx h$

$\Rightarrow \mathrm{\Delta}{p}_{y}\mathrm{\Delta}y\ge \frac{h}{2\pi}$

$\Rightarrow \mathrm{\Delta}p\mathrm{\Delta}x\ge \frac{h}{2\pi}$

asked 2022-09-25

If matter has a wave nature, why is this wave - like characteristic not observable in our daily experiences?