. Which Of The Following Will Produce Diffuse Reflection Of Light?
2 answers:
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Answer:
Explanation:
Radius of dee, r = 8 mm = 0.008 m
Electric field, e = 400 V/m
Magnetic field, B = 4.7 x 10^-4 T
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
(a) Let v is the speed of electrons.
v = 661098.9 = 661099 m/s
(b)
e / m = 1.76 x 10^14 C / kg
(c) Let K be the kinetic energy
K = 0.5 x mv²
K = 0.5 x 9.1 x 10^-31 x 661099 x 661099
K = 1.99 x 10^-19 J
K = 1.24 eV
So, the potential difference is
V = 1.24 V
(d) if the acceleration voltage is doubled
V = 2 x 1.24 = 2.48 V
So, Kinetic energy
K = 2.48 eV
K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J
Let v is the speed
K = 0.5 x mv²
3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²
v = 933856.5 m/s
Let the new radius is r.
r = 0.0113 m = 1.13 cm
Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ =
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV