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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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<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

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Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

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So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

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