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2 years ago

Through what potential difference must an electron be accelerated from rest to have a de broglie wavelength of 400 nm ?

Physics

1 answer:

FromTheMoon [43]2 years ago

7 0

So the potential difference is 9.4229μV

Given,

mass of the electron = 9.1*10^{-31}kg

charge of electron = 1.6*10^{-19}C

From Broglie's relation we can write

mv=h/λ

So, v=h/mλ= $\frac{6.626*10^{-34} }{(9.1*10^{-31}*(400*10^{-9} ) )}$ =1820.32m/s

Kinetic Energy of electron = $\frac{1}{2} mv^{2}$ =eV

V= $\frac{1*m*v^{2} }{2*e}$ = $\frac{1*9.1*10^{-31}*1820.32^{2} }{2*1.6*10^{-19} }$ =9.4229μV

<h3>Electron</h3>

Unattached or attached to an atom, an electron is a negatively charged subatomic particle (not bound). One of the three main types of particles inside an atom, along with protons and neutrons, is an electron that is linked to the atom.

An atom's nucleus is made up of electrons, protons, and neutrons. The negative charge of the electron is balanced by the positive charge of the proton. An atom is in a neutral state if its protons and electrons are equal in number.

The differences between electrons and the other particles are numerous. They have a mass that is considerably lower, exist outside of the nucleus, and display both wave- and particle-like properties.

Learn more about Electron here:

brainly.com/question/1255220

#SPJ4

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Which phenomena did Michael Faraday explore to help him invent the motor and the generator?

mamaluj [8]

The best and most correct answer among the choices provided by the question is the fourth choice. The phenomenon that Michael Faraday explored was electricity and magnetism. I hope my answer has come to your help. God bless and have a nice day ahead!

8 0

3 years ago

Read 2 more answers

Suppose that a nascar race car is moving to the right with a constant velocity of +86 m/s. What is the average acceleration of t

Stels [109]

(a) As the car is moving with constant velocity, it means the rate change of velocity does not change, therefore the average acceleration of the car is zero.

Thus, there is no acceleration, when velocity is constant.

(b) Average acceleration,

$a =\frac{ v-u}{\Delta t}$

Here, v is final velocity and u is the initial velocity and $\Delta t$ is the time interval.

As twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed, therefore

$a =\frac{ -86 \ m/s-86 \ m/s }{12 \ s} = - 14 .3 \ m/s^2$

Thus, the average acceleration of the car is $14 .3 \ m/s^2$ in the direction to the left.

6 0

3 years ago

Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1

harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum $p_{i}$ (before the elastic collision) must be equal to the final momentum $p_{f}$ (after the elastic collision):

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{1}V_{i} + m_{2}U_{i}$

$p_{f}=m_{1}V_{f} + m_{2}U_{f}$

Where:

$m_{1}=200 kg +100 kg=300 kg$ is the combined mass of Tubby and Libby with the car

$V_{i}=10 m/s$ is the velocity of Tubby and Libby with the car before the collision

$m_{2}=25 kg + 100 kg=125 kg$ is the combined mass of Flubby with its car

$U_{i}=0 m/s$ is the velocity of Flubby with the car before the collision

$V_{f}=4.12 m/s$ is the velocity of Tubby and Libby with the car after the collision

$U_{f}$ is the velocity of Flubby with the car after the collision

So, we have the following:

$m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f}$ (2)

Finding $U_{f}$ :

$U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}}$ (3)

$U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg}$ (4)

Finally:

$U_{f}=14.1 m/s$

8 0

3 years ago

What is formed when two surfaces come into contact

balu736 [363]

Friction is Produced there.....

7 0

3 years ago

When a mass of 29 g is attached to a certain spring, it makes 20 complete vibrations in 3.1 s. What is the spring constant of th

never [62]

Answer:

The spring constant of the spring is 47.62 N/m

Explanation:

Given that,

Mass that is attached with the spring, m = 29 g = 0.029 kg

The spring makes 20 complete vibrations in 3.1 s. We need to find the spring constant of the spring. We know that the number of oscillations per unit time is called frequency of an object. So,

$f=\dfrac{20}{3.1}$