Question

If an electron's position can be measured to a precision of 15nm, what is the uncertainty in its speed? Assuming the minimum speed must be at least equal to its uncertainty, what is the electron's minimum kinetic energy (in eV)?

Answer 1

Answer:3867.2 m/s

Step-by-step explanation:

The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called wave–particle duality. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle.

λ=hmv(6.4.3)

The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the fundamental vibration, and higher-energy vibrations are overtones and have successively more nodes, points where the amplitude of the wave is always zero. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.

Answer 2

3867.2 m/s

4.25 x 10⁻⁵ eV

Step-by-step explanation:

Δx = uncertainty in electron's position = 15 nm = 15 x 10⁻⁹ m

Δv = uncertainty in the velocity of electron = ?

m = mass of electron = 9.1 x 10⁻³¹ kg

Using heisenberg uncertainty principle

`m \Delta x \Delta v = (h)/(4\pi )`

`(9.1* 10^(-31)) (15* 10^(-9)) \Delta v = (6.63* 10^(-34))/(4(3.14) )`

Δv = 3867.2 m/s

v = minimum speed of electron = 3867.2 m/s

Minimum kinetic energy is given as

KE = (0.5) m v²

KE = (0.5) (9.1 x 10⁻³¹) (3867.2)²

KE = 6.8 x 10⁻²⁴ J

KE =
`(6.8* 10^(-24))/(1.6* 10^(-19))eV`

KE = 4.25 x 10⁻⁵ eV