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Calculate the de-Broglie wavelength of an electron acelerated from ...


What is the de Broglie wavelength associated with an electron that's accelerated through a potential difference of 100 volts? the following answer which has ...


For more information, see Calculate the de-Broglie wavelength of an electron acelerated from ...

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quantum mechanics | Definition, Development, & Equations ...


Author of Problems in Quantum Mechanics with Solutions and others. ... The energy radiated at different wavelengths is a maximum at a wavelength that ... to the energy difference between the initial and final states in a transition. Prior to ... of the waves is related to the momentum of the electrons by the de Broglie equation.


For more information, see quantum mechanics | Definition, Development, & Equations ...

Qalaxia Knowlege Bot
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I found an answer from en.wikipedia.org

Transmission electron microscopy - Wikipedia


The electrons are then accelerated by an electric potential (measured in volts) and focused by electrostatic and electromagnetic lenses onto the sample.


For more information, see Transmission electron microscopy - Wikipedia

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I found an answer from opentextbc.ca

De Broglie's Matter Waves – University Physics Volume 3


Compton's formula established that an electromagnetic wave can behave like a particle of ... Today, this idea is known as de Broglie's hypothesis of matter waves . ... of a new theory of wave quantum mechanics to describe the physics of atoms and ... from de Broglie relations that matter waves satisfy the following relation:.


For more information, see De Broglie's Matter Waves – University Physics Volume 3

Swetha
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Given that

Magnitude of accelerating potential of an electron V = 100 volts

Wavelength associated with an electron = ?


Step 1: Plug in the known values in the de Broglie wavelength

                 Broglie wavelength of the electron

                             \lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}

                                        or

                              \lambda = \frac{1.227}{\sqrt{V}}

                          Where,  V - magnitude of accelerating potential in volts


                             \lambda = \frac{1.227}{\sqrt{100}}

                             \lambda = 0.123 nm


In this case, the de Broglie wavelength correlated with the electron is of

X-ray wavelength order