I found an answer from www.quora.com

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 ...

I found an answer from www.britannica.com

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 ...

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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**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

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