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Margarita [4]
2 years ago
7

Can an electron with a de broglie wavelength of 2 μm pass through a slit that is 1 μm wide? select the correct answer and explan

ation
Physics
1 answer:
borishaifa [10]2 years ago
3 0

No electron with a de Broglie wavelength of 2 μm can not pass through a slit that is 1 μm wide

When studying quantum mechanics, the de Broglie wavelength is a key idea. De Broglie wavelength is the wavelength () that is connected to an item in relation to its momentum and mass. Typically, a particle's force is inversely proportional to its de Broglie wavelength.

Where "h" is the Plank constant, momentum has the formula = h m v = h. The de Broglie equation and de Broglie wavelength are terms used to describe the relationship between a particle's momentum and wavelength. The probability density of locating an object at a specific location in the configuration space is determined by the De Broglie wavelength, which is a wavelength present in all quantum mechanical objects. A particle's momentum and de Broglie wavelength are inversely related.

To learn more about de Broglie wavelength please visit-
brainly.com/question/17295250
#SPJ4

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A watermelon is dropped from the top of a 80m tall building. We want to find the velocity of the watermelon after it falls for 1
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Answer:

v_f = v_i + at

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now if water melon start from rest then we have

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acceleration due to gravity for watermelon

a = 9.81 m/s^2

now we need to find the final speed of watermelon

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7 0
4 years ago
During which season of the year will your nails grow most quickly?
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in positive numbers less than 1, the zeros between the decimal point and a non-zero number are _______ significant?
DedPeter [7]

Answer:

Explanation:

If a number of less than 1, then the number has a decimal point like

0.085, 0.008 e.t.c.

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But if there a zero between the none zero e.g. 0.0087056

Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6

But if the zero is not in between the none zero digit, then the zero is insignificant

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8 0
4 years ago
2. A certain object revolves at a rate of 30 rpm. Please determine the frequency and period of this
schepotkina [342]

Answer:

The time period of the motion is, T = 0.03 s

The frequency of the rotation is, f = 30 Hz

Explanation:

Given data,

The rotational speed of an object, ω = 30 rpm

                                                          ω = 188.5 rad/s

The time period of motion is,

                            T = 2π / ω

Substituting the given values in the above equation

                               = 2π / 188.5

                            T = 0.03 s

The time period of the motion is, T = 0.03 s

The frequency of rotation,

                                f = 1 /T

                                   = 1 / 0.03

                                   = 30 Hz

Hence, the frequency of the rotation is, f = 30 Hz

7 0
4 years ago
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