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An **electron**, an alpha **particle**, and a **proton have** the **same kinetic** ...

An **electron**, an alpha **particle**, and a **proton have** the **same kinetic energy**. Which
of **these particles has** the **shortest de**-**Broglie wavelength**? 7 Answers.

For more information, see An **electron**, an alpha **particle**, and a **proton have** the **same kinetic** ...

I found an answer from en.wikipedia.org

**Proton** - Wikipedia

A **proton is** a subatomic **particle**, symbol p or p ^{+} , with a positive electric charge
of +1e ... In previous years, Rutherford **had** discovered that the hydrogen nucleus
(known ... **particles**, in the modern Standard Model of **particle** physics, **protons** are
... As a muon **is** 200 times heavier than an **electron**, its **de Broglie wavelength is** ...

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

I found an answer from www.britannica.com

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

**These** properties include the interactions of the **particles** with one another and ...
**is** converted into the **kinetic energy** meu^{2}/2 of the emitted **electron** (me **is** the ...
He found that part of the scattered radiation **had** the **same wavelength** λ0 as the
... radiation **has** both **particle** and wave characteristics, Louis-Victor **de Broglie** of
...

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

I found an answer from opentextbc.ca

**De Broglie's Matter Waves** – University **Physics** Volume 3

Today, this idea is known as **de Broglie's** hypothesis of **matter waves**. In 1926, **De**
**Broglie's** hypothesis, together with Bohr's early quantum **theory**, ... Any **particle**
that has **energy** and momentum is a **de Broglie wave** of frequency f and
**wavelength** \lambda : ... For **matter waves**, this group velocity is the velocity u of
the **particle**.

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

Given that

Kinetic energy of electron = Kinetic energy of alpha-particles = Kinetic energy of proton.

Step 1: Set up an equation for the kinetic energy in terms of 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}

Step 2: Get an expression for relation between the de Broglie wavelength and mass

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{h}{\sqrt{2mK.E}}

\lambda \propto \frac{1}{\sqrt{m}}

For the same Kinetic-energy, the de Broglie wavelength associated with the particle is inversely proportional to the square

root of its mass. Thus, the higher mass particle has the shortest wavelength.

In this case, the protons are heavier than the electrons, and the alpha particles four times that of a proton. The alpha

particle is also the heaviest compared to the proton and the electron. thus alpha-particle has the shortest de Broglie

wavelength.