## How do you calculate the de Broglie wavelength?

Apply the de Broglie wave equation λ=hmv λ = h m v to solve for the wavelength of the moving electron. Step 3: Think about your result. This very small wavelength is about 1/20th of the diameter of a hydrogen atom. Looking at the equation, as the speed of the electron decreases, its wavelength increases.

**Which of the following particle moving with same velocity has maximum de Broglie wavelength?**

Hence, the particle with the maximum wavelength is the electron.

**Is the following matter waves travel with equal velocity the longest wavelength is that of?**

If moving with equal speeds, the longest wavelength of the following matter waves is that for a(an) a)electron b)alpha particle c)proton d)neutron please answer fast. The answer is electron with a wavelength of 1.23nm.

### Which of the following gives the de Broglie relationship?

p=mvh.

**What is the velocity of an electron that has a de Broglie wavelength of 1 cm?**

The speed of this electron is equal to 1 c divided by 100, or 299,792,458 m/s / 100 = 2,997,924.58 m/s .

**What is the de Broglie wavelength of an electron when moving with the velocity of 0.1 c?**

Answer. So λ=h/(mv), where h is Planck’s constant, m is the mass of an electron, and v = 0.1c.

43×10’4m.

**What is the de Broglie wavelength of an electron whose kinetic energy is 100 eV?**

0.123 nm

**What is de Broglie wavelength of an electron whose kinetic energy is 120 eV?**

Thus wavelength of the given electron of kinetic energy of 120 eV is 4.5 ×10^-20 m.

## What is the de Broglie wavelength of an electron that has been accelerated?

1226 nm.

**What is the wavelength associated with 200 eV electron?**

m = 0.0275 nm.

**What will be the wavelength of the electron of energy 100 eV?**

we get λ=1.

### What is the momentum of electron of energy 100 eV?

Electrons in the atoms exist in spherical shells of various radii, that represent the energy levels. Therefore, the momentum of an electron of energy 100 eV is 1.227 × 10^{-10} m.

**What is the momentum speed and de Broglie wavelength of an electron with kinetic energy of 120 eV?**

(c) de Broglie wavelength of an electron with kinetic energy of 120 eV. Therefore, the momentum of the electron is 5.91 × 10’24 kg m s’1 .

**What is the kinetic energy of a neutron?**

Therefore, the kinetic energy of the neutron is 6.75 × 10 ’21 J or 4.219 × 10 ‘2 eV.

The relationship between momentum and wavelength for matter waves is given by p = h/λ, and the relationship energy and frequency is E = hf. The wavelength λ = h/p is called the de Broglie wavelength, and the relations λ = h/p and f = E/h are called the de Broglie relations.

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