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1 year ago

What is the de broglie wavelength of an electron (m = 9. 11 × 10-31 kg) moving at a velocity of 3. 0 × 107 m/s?

Chemistry

1 answer:

SIZIF [17.4K]1 year ago

6 0

the de Broglie wavelength of the electron moving at a velocity of 3. 0 × 107 m/s is 2.4×10∧-11nm

lambda = h/mv

The de Broglie wavelength is defined as follows: lambda = h/mv, wherein the greek letter lambda represents the wavelength, h is Planck's consistent, m is the particle's mass and v is its pace. One may also explicit mv as the particle's momentum

according to wave-particle duality, the De Broglie wavelength is a wavelength manifested in all the gadgets in quantum mechanics which determines the chance density of locating the item at a given factor of the configuration area. The de Broglie wavelength of a particle is inversely proportional to its momentum

Learn more about de Broglie wavelength here: brainly.com/question/19052579

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Answer:

Elements that belong to the <em>same </em><em>GROUP</em><em> </em>of the periodic table have the most similar chemical properties.

Explanation:

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3 years ago

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Answer:

<h2>9.00 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

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N is the number of entities

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From the question we have

$n = \frac{5.42 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 9.0033$

We have the final answer as

<h3>9.00 moles</h3>

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