If the de Broglie wavelength of an electron is 2.4 μm,what is the speed of the electron? $\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.$ $h = 6.626 \times 10 ^{- 34} \mathrm {~J} \cdot \mathrm { s } )$ A)3.0 × 10² m/s B)2.5 × 10⁵ m/s C)1.7 × 10³ m/s D)8.3 × 10⁶ m/s

Question

Quizplus · Accepted Answer

The de Broglie wavelength of an electron is given by the equation:

λ = h/p

where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the electron. We can rearrange this equation to solve for the momentum:

p = h/λ

Once we know the momentum, we can use the equation for the momentum of a particle:

p = mv

where m is the mass of the electron, v is the velocity of the electron, and p is the momentum. We can rearrange this equation to solve for the velocity:

v = p/m

Substituting the values given in the problem, we get:

p = h/λ = (6.626 x 10^-34 J.s)/(2.4 x 10^-6 m) = 2.77 x 10^-28 kg.m/s

v = p/m = (2.77 x 10^-28 kg.m/s)/(9.11 x 10^-31 kg) = 3.04 x 10^7 m/s

Therefore, the answer is A: 3.0 x 10^7 m/s.

If the De Broglie Wavelength of an Electron Is 2 (melectron =9.11×10−31 kg,\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.(melectron ​=9.11×10−31 kg,

If the De Broglie Wavelength of an Electron Is 2 $\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.$