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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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HELP ME the bending of light when it changes media is called _________ u can get brainliest

olga55 [171]

Answer:

Refraction

Explanation:

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3 years ago

Read 2 more answers

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

I hope it helps you!

3 0

3 years ago

If the Sun were scaled down to the size of a grapefruit, about how far would you have to walk from our classroom to reach Alpha

Burka [1]

Answer:

2000 miles.

Explanation:

It's the Colorado scale model, in which sun is taken as the grapefruit and the distances are measured for different planets with respect to sun just for understanding.

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3 years ago

How much time will it take for a coyote to travel 48 meters across the field to get to the unsuspecting rabbit eating grass in t

kkurt [141]

Formula for time
t=d/s
so…
t= 48m/4m/s
the two ms cancel each other out and ur left with s

t=12s

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3 years ago

The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO.

Genrish500 [490]

Answer:

Explanation:

Given a square side loop of length 10cm

L=10cm=0.1m

Then, Area=L²

Area=0.1²

Area=0.01m²

Given that, frequency=60Hz

And magnetic field B=0.8T

a. Flux Φ

Flux is given as

Φ=BA Sin(wt)

w=2πf

Φ=BA Sin(2πft)

Φ=0.8×0.01 Sin(2×π×60t)

Φ=0.008Sin(120πt) Weber

b. EMF in loop

Emf is given as

EMF= -N dΦ/dt

Where N is number of turns

Φ=0.008Sin(120πt)

dΦ/dt= 0.008×120Cos(120πt)

dΦ/dt= 0.96Cos(120πt)

Emf=-NdΦ/dt

Emf=-0.96NCos(120πt). Volts

c. Current induced for a resistance of 1ohms

From ohms law, V=iR

Therefore, Emf=iR

i=EMF/R

i=-0.96NCos(120πt) / 1

i=-0.96NCos(120πt) Ampere

d. Power delivered to the loop

Power is given as

P=IV

P=-0.96NCos(120πt)•-0.96NCos(120πt)

P=0.92N²Cos²(120πt) Watt

e. Torque

Torque is given as

τ=iL²B

τ=-0.96NCos(120πt)•0.1²×0.8

τ=-0.00768NCos(120πt) Nm

8 0

2 years ago