What would be the de Broglie wavelength for an electron (m=1x10^-31kg) moving at 2.4x10^6 m/s ?

Answer 1

and a very merry xmas to you to

Answer 2

The de broglie wavelength can be calculated using the formula

λ = h/mv where h = Planck's constant

λ= h/mv
= 6.626*10^-34 / ( 1*10^-31 * 2.4 * 10^6 )
= 2.76 * 10^-9 m

Ans: The De Broglie Wavelength of the electron is 2.76 *10^-9 m or 2.76 nanometres


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Answer 3

if a electron was moving freely it would oscilate with an energy equal to 1.6 x10^-19 joules it wavelenght per cycle would be its diameter multiplied by its frequency.

If the electron is locked in to the atoms it will perfom an oscilation nested within atnother oscilation(De Broglie frequency)
Knowing the diameter and velocityof the electron the different oscilations of the electron can be calculated.

DeBroglie calculated the transerse oscillation using The energy time of the atom as a constant. In the Case of the Hydrogen atom the Electron traverse wavelenght
oscilation is =h/m x v , where h is Planck's constant ,m is electron electric mass and v is the orbital velocity of the electron.
De Broglies assumed that the mass of an electron is constant.
However as per Relativity theory any moving mass traveling at relativistic speeds changes as a function of velocity.

Answer 4

de Broglie wavelength is given by
L = h/mv
where h is the Planck's constant, m is the mass of the electron and v the velocity of the electron.
L = 6.63 x 10^-34/(9.1 x 10^-31 x 2.4 x 10^6)
= 304.17 m
Mass of the electron is not 1 x 10^-31 kg. It is 9.1 x 10^-31 kg. You can approximate it to 1 x 10^-30 kg

Answer 5

f = h/mv , and h =6.626 * 10^(-34) Planck's constant

f= h/mv
= 6.626 * 10^(-34) / ( 1*10^-31 * 2.4 * 10^6 )
= 2.7608 * 10^(-9) m
~= 2.76nm