I found an answer from www.quora.com

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

According to Louis **de Broglie's** famous **equation**, Wave Nature (**Wavelength**) ...
**Momentum** = sqrt(2 x Mass x Kinetic Energy) [ As K.E. = 0.5mv^2, and ... The
question isn't quite right, **potential difference** is specified in **V** (**volt**) not eV (
**electron volt**). ... If an **electron** is **accelerated** from rest **through** a **potential**
**difference** of 10V, ...

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

I found an answer from en.wikipedia.org

**Electron** - Wikipedia

**De Broglie's** prediction of a wave nature for **electrons** led Erwin SchrÃ¶dinger to
postulate a wave **equation** for **electrons** moving under the influence of the
nucleus ...

For more information, see **Electron** - Wikipedia

I found an answer from www.khanacademy.org

**De Broglie wavelength** (video) | Khan Academy

Bohr model **energy** levels (derivation using **physics**) ... Since **electrons** have a
rest mass, unlike **photons**, they have a **de Broglie wavelength** ... This means that
a microscope using **electron** "**matter waves**" instead of **photon light waves** can
see ... For example in the **double** slit experiment.... **electron waves** diffract and
then ...

For more information, see **De Broglie wavelength** (video) | Khan Academy

Given that

Potential difference(voltage) V = 56 V

a) Calculate the momentum

Recall the formula of momentum and kinetic energy

Momentum p = mv

Kinetic energy K.E = \frac{1}{2} mv^2

Where, p - momentum, m - mass and v - velocity

K.E = \frac{1}{2} m * (\frac{p}{m})^2 \because p=mv

K.E = \frac{p^2}{2m}

p = \sqrt{2mK.E}

According to the conversion law

K.E = eV

Where, charge of electron e = 1.6* 10^{-19} and V - potential difference(voltage)

Therefore, momentum p = \sqrt{2meV}

Substituting the known values in the above equation

Mass of an electron = 9.1 * 10^{-31} kgs

p = \sqrt{2 * 9.1 * 10^{-31} * 1.6*10^{-19} V * 56}

p = 4.03 *10^{-24} kg m/s

Hence, Momentum p = 4.03 *10^{-24} kg m/s

b) de Broglie wavelength of an electron

Recall the de Broglie wavelength formula

de Broglie wavelength \lambda = \frac{h}{p}

Where, h - Planck constant = 6.63* 10^{-34} J and p - linear momentum.

\lambda = \frac{6.63*10^{-34}}{4.03*10^{-24}}

\lambda = 1.64*10^{-10} m

\lambda = 0.164 nm

Therefore, de Broglie wavelength of an electron \lambda = 0.164 nm