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hoa [83]
1 year ago
6

What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0. 85 nm?

Physics
1 answer:
Alex_Xolod [135]1 year ago
4 0

Speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0. 85 nm is 8.6*10^5m/s.

To find the answer, we have to know about the energy of photon.

<h3>What is the speed of the electron here?</h3>
  • As we know that the momentum of an x-ray photon with a wavelength w,

                            P=\frac{h}{w}

where; h is the plank's constant.

  • Thus, the momentum will be,

            P=\frac{6.63*10^{-34}}{0.85*10^{-9}} =7.8*10^{-25}kgm/s.

  • We have to find the speed of the electron, thus, we have the expression of linear momentum as,

                          P=mv\\v=\frac{P}{m} =\frac{7.8*10^{-25}}{9.1*10^{-31}}=8.6*10^5m/s.

Thus, we can conclude that, speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0. 85 nm is 8.6*10^5m/s.

Learn more about the energy of photon here:

brainly.com/question/3584036

#SPJ4

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VladimirAG [237]

Answer:

4.9612 s

Explanation:

Applying,

T = 2π√(L/g)............... Equation 1

Where T = period of the pendulum, L = Lenght of the pendulum, g = acceleration due to gravity of the moon, π = pie.

From the question,

Given: L = 1 m, g = 1.6 m/s²

Constant: π = 3.14

Substitute these values into equation 1

T = 2×3.14×√(1/1.6)

T = 6.28√(0.625)

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6 0
3 years ago
A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretch
r-ruslan [8.4K]

Answer:

a) A = 0.98 m

b) Ф = 90°

c) x = -0.98sin(12.25t)

Explanation:

We know the value of the spring constant which is 300 N/m, the innitial apmplitude, that we will call it xo is 0, at t = 0, and the speed is 12 m/s

The expression for the amplitude under these conditions is:

A = √xo² + vx²/w² (1)

To calculate the angular speed w, we use the following expression:

w = √k/m  (2)

Calculating w:

w = √300/2 = 12.25 rad/s

Now, we replace this value into equation 1, along with the other known values and solve for A:

A = √0 + (12)²/(12.25)²

A = 0.98 m

b) In this part, is actually easy, the displacement of x in function of the time is given by:

x = A cos(wt - Ф) (4)

But at t = 0 we have then:

x = xo = A cosФ (5)

Solving for the angle Ф we have:

xo/A = cosФ

Ф = arccos(x0/A)  (6)

Replacing the data in (6):

Ф = arccos(0/0.98)

Ф = 90°

c) Equation (4) is the expression for the simple harmonic motion

x = A cos(wt - Ф)

And if we replace here the value of w and the previous angle, we can write an equation for x in function of t:

x = 0.98 cos (12.25t - 90)

x = 0.98 cos(12.25t - 90)

And we have an trigonometric expression for cos that is:

cos(α - π/2) = -sinα

in this case, α will be the value of w = 12.25 rad/s. and 90° is the same as rewritting as π/2, therefore:

cos(α - π/2) = cos(12.25t - 90)

x = -0.98sin(12.25t)

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Likurg_2 [28]
Hey there!

The answer is : 
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</span>
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