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?

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

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

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photocells convert light energy to electricity