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Varvara68 [4.7K]
3 years ago
15

Violet light of wavelength 405 nm ejects electrons with a maximum kinetic energy of 0.890 eV from a certain metal. What is the b

inding energy (in electronvolts) of electrons to this metal
Physics
1 answer:
AleksandrR [38]3 years ago
4 0

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 = \frac{hc}{\lambda } - K

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

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The second arrow from the top to the bottom

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This can be explained by the Newton's third law.

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When an object A exert a force to an object B. Then object B also excerts a force with the same size but opposite direction to object A.

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A trough is 3 meters long, 2 meters wide, and 4 meters deep. The vertical cross-section of the trough parallel to an end is shap
irina1246 [14]

Answer:

work done is 470400 J

Explanation:

given data

long = 3 m

wide = 2 m

deep = 4 m

density = 1000 kg/m³

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Find the amount of work in joules

solution

we know here volume = area ×  length

so

area = 0.5 ×base×height

area = 0.5 ×2×4

area = 4 m²

so volume = 4 × 3 = 12 m³

and

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8 0
3 years ago
A solid sphere of radius 30cm is uniformly charged to 100nC. a) What is the volume charge density of the sphere? b) What is the
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Answer

given,

total charge Q = 100 n C

                        = 100 × 10⁻⁹ C

radius of the solid sphere = 30 cm

                                           = 0.3 m

Volume of sphere = \dfrac{4}{3}\pi r^3

                              = \dfrac{4}{3}\pi\times 0.3^3

                              =0.113 m³

a) volume charge density

\rho = \dfrac{10^{-7}}{0.133}

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b) at r = 10 cm = 0.1 m

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Q = \dfrac{4}{3}\pi\times 0.1^3\time \rho

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E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}

E_1 = \dfrac{3.7037 \times 10^{-9}}{4\pi \times 8.85\times 10^{-12}\times 0.1^2}

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at r = 20 cm = 0.2 m

Q = \dfrac{4}{3}\pi\times r^3\time \rho

E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}

E_1 = \dfrac{ \dfrac{4}{3}\pi\times r^3\time \rho}{4\pi \epsilon_0 r^2}

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E_1 = \dfrac{\rho}{3 \epsilon_0}

E_1 = \dfrac{0.3\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}

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