Write the mathemetical relation between work force and displacement

Physics

1 answer:

il63 [147K]2 years ago

8 0

Answer:

Work is measured as the product of force and the displacement in the direction of the force. Work = force × displacement in the direction of the force.

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A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a s

tatiyna

Answer:

a. λ = 647.2 nm

b. I₀ 9.36 x 10⁻⁵

Explanation:

Given:

β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m

The wavelength of the radiation can be find using

β = 2 π / γ * sin θ

λ = [ 2π * γ * sin θ ] / β

λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad

λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm

The intensity of the central maximum I₀

I = I₀ (4 / β² ) * sin ( β / 2)²

I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²

I = I₀ 9.36 x 10⁻⁵

8 0

3 years ago

Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to

jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.