Question

Consider electrons incident on a double slit apparatus. You observe that they form the same pattern as 465-nm light on the screen behind the slits atan) - 25% Part (a) Calculate the velocity, in meters per second, of the electrons Grade Deduct Potenti Submi Attemp (09% pe tan() π ( cosO cotan() asin()acos() atanO acotanO sinh cosh)tanh)cotanh) Degrees Radians sin -- detaile 0 END BACKSPACE DEL CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 2 Feedback: 2% deduction per feedback. 25% Part (b) Calculate the energy, in joules, of the a photon with this wavelength Δ 25% Part (c) Calculate the kinetic energy, in joules, of the electron 25% Part (d) How many times greater is the energy of the photon than the electron?

Answer 1

a) v = 1.567 10³ m / s , b) E = 4,278 10⁻¹⁹ J , c) E / K = 3.8 10⁵

Step-by-step explanation:

a) For this exercise we will use the d’Broglie duality principle, which relates

p = h / λ

Where p is the moment, h the Planck constant that is worth .6.63 10⁻³⁴ J s and λ the wavelength

The momentum of a particle is defined by

p = mv

m v = h / λ

v = h / m λ

Reduce to SI units

λ = 465 nm (1m / 10⁹ nm) = 465 10⁻⁹ m

v = 6.63 10⁻³⁴ / (9.1 10⁻³¹ 465 10⁻⁹)

v = 1.567 10³ m / s

b) The Planck equation is

E = h f

The speed of light is

c = λ f

Let's replace

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/465 10⁻⁹

E = 4,278 10⁻¹⁹ J

c) The kinetic energy is

K = ½ m v²

K = ½ 9.1 10⁻³¹ (1.567 10³) 2

K = 11.17 10⁻²⁵ J

d) the energy ratio is

E / K = 4,278 10⁻¹⁹ / 1,117 10⁻²⁴

E / K = 3.8 10⁵