Ask question

Ask question

All categories

2 years ago

12

Wave-particle duality tells us that wave and particle models apply to all objects whatever the size, so why don't we observe wav

e properties in macroscopic objects?

1 answer:

Genrish500 [490]2 years ago

4 0

Answer:

Because the wavelengths of macroscopic objects are too short for them to be detectable.

Explanation:

Wavelength of an object is given by de Broglie wavelength as:

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

Where, 'h' is Planck's constant, 'm' is mass of object and 'v' is its velocity.

So, for macroscopic objects, the mass is very large compared to microscopic objects. As we can observe from the above formula, there is an inverse relationship between the mass and wavelength of the object.

So, for vary larger masses, the wavelength would be too short and one will find it undetectable. Therefore, we don't observe wave properties in macroscopic objects.

You might be interested in

Most binary systems with an invisible companion contain a large, bright star and a small, dim star hidden by the light of its la

Vlada [557]

Answer:

Brain signals are converted

8 0

2 years ago

An infinite charged wire with charge per unit length lambda lies along the central axis of a cylindrical surface of radius r and

serg [7]

<h2>The flux through the infinite charged wire along the central axis of a cylindrical surface of radius r and length l is ∅E = E x 2πrl </h2>

Explanation:

let us consider a thin infinitely long straight wire having a uniform charge density λ Cm⁻¹.To determine the field at a distance r from the line charge , we have cylindrical gaussian surface of radius r, length l,and with its axis along the line charge. it has curved surface S₁ , and flat circular ends S₂ and S₃. Obviously, dS₁//E, dS₂ ⊥E , and dS₃ ⊥ E , so, only the curved surface contributes towards the total flux.

∅E = ∫ E.dS = ∫E.dS₁ +∫E.dS₂ +∫E.dS₃

= ∫EdS₁ cos0⁰ +∫EdS₂ cos 90⁰ +∫Eds₃ cos 90⁰

= E∫ds₁₁ +0+0

= E x area of curved surface

∅E = E x 2πrl

3 0

2 years ago

Use Wien’s Law to calculate the peak wavelength of Betelgeuse, based on the temperature found in Question #8. Note: 1 nanometer

kodGreya [7K]

The peak wavelength of Betelgeuse is 828 nm

Explanation:

The relationship between surface temperature and peak wavelength of a star is given by Wien's displacement law:

$\lambda=\frac{b}{T}$

where

$\lambda$ is the peak wavelength

T is the surface temperature

$b=2.898\cdot 10^{-3} m\cdot K$ is Wien's constant

For Betelgeuse, the surface temperature is approximately

T = 3500 K

Therefore, its peak wavelength is:

$\lambda=\frac{2.898\cdot 10^{-3}}{3500}=8.28\cdot 10^{-7} m = 828 nm$

Learn more about wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

8 0

2 years ago

Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellit

GrogVix [38]

Answer:

1) C

2) E

Explanation:

3 0

2 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago