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Misha Larkins [42]
3 years ago
9

How does the de broglie wavelength of an electron change if its momentum increases?

Physics
1 answer:
Virty [35]3 years ago
3 0

Answer:

The De Broglie wavelength decreases

Explanation:

The relationship between the De Broglie wavelength of a particle and its momentum is given by

\lambda=\frac{h}{p}

where

\lambda is the De Broglie wavelength of the particle

h is the Planck constant

p is the momentum of the particle

As we see from the formula, there is an inverse relationship between the De Broglie's wavelength and the momentum. Therefore, we can conclude that:

- if the momentum of the electron increases,

- its De Broglie wavelength will decrease

and vice-versa.

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now here we know that

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the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f
vesna_86 [32]

Answer:

6.63\times 10^8\ N/C

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

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\dfrac{E}{B}=c\\\\E=Bc

Put all the values,

E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C

So, the magnitude of the electric field is equal to 6.63\times 10^8\ N/C.

7 0
3 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
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