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What is the De Broglie wavelength of an electron under 150 V acceleration?

yanalaym [24]

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron $m=9.1\times 10^{-31}$

Let the velocity of electron = v

Charge on the electron $=1.6\times 10^{-19}C$

plank's constant h = $6.67\times 10^{-34}$

According to energy conservation $eV =\frac{1}{2}mv^2$

$v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec$

Now we know that De Broglie wavelength $\lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm$

4 0

3 years ago

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An engineer designing a hydroelectric facility recalls that the flow rate (in gallons per second) is four times the value of the

GarryVolchara [31]

Answer:

Flow rate = 86.48 gal/s

Head of water = 21.62 ft

Explanation:

Detailed explanation and calculation is shown in the image below.

7 0

4 years ago

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What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00

yKpoI14uk [10]

Answer:

The velocity of flow is 10.0 m/s.

Explanation:

We shall use Manning's equation to calculate the velocity of flow

Velocity of flow by manning's equation is given by

$V=\frac{1}{n}R^{2/3}S^{1/2}$

where

n = manning's roughness coefficient

R = hydraulic radius

S = bed slope of the channel

We know that for an asphalt channel value of manning's roughness coefficient = 0.016

Applying values in the above equation we obtain velocity of flow as

$V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s$

7 0

3 years ago

Find an expectation value in the n th state of the harmonic oscillator.Find an expectation value in the n th state of the harmon

s344n2d4d5 [400]

The classical motion for an oscillator that starts from rest at location x₀ is

x(t) = x₀ cos(ωt)

The probability that the particle is at a particular x at a particular time t

is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average

to get the spatial density. Our natural time scale for the averaging is a half

cycle, take t = 0 → π/ ω

Thus,

ρ = $\frac{1}{\pi / w} \int\limits^\pi_0 {d(x - x_o cos(wt))} \, dt$

Limit is 0 to π/ω

We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt) so that

ρ(x) = $-\frac{w}{\pi } \int\limits^x_x {\frac{d ( x - y)}{x_ow sin(wt)} } \, dy$

Limit is x₀ to -x₀

$\frac{1}{\pi } \int\limits^x_x {\frac{d (x-y)}{x_o\sqrt{1 - cos^2(wt)} } } \, dy$

Limit is -x₀ to x₀

$= \frac{1}{\pi } \int\limits^x_x {\frac{d(x-y)}{\sqrt{x_o^2 - y^2} } } \, dy\\ \\= \frac{1}{\pi\sqrt{x_o^2 - x^2} }$

This has $\int\limits^x_x {p(x)} \, dx = 1$ as expected. Here the limit is -x₀ to x₀

The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.

6 0

4 years ago

The Leaning Tower of Pisa is located in which country? Question 1 options: France Italy The United States Norway

Ket [755]

Answer:

Tuscany, Central Italy

Explanation:

5 0

3 years ago

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