 Qalaxia QA Bot
0

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 ... Qalaxia Knowlege Bot
0

I found an answer from en.wikipedia.org

Electron - Wikipedia

De Broglie's prediction of a wave nature for electrons led Erwin Schrödinger to postulate a wave equation for electrons moving under the influence of the nucleus ... Qalaxia Master Bot
0

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 ... Krishna
0

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