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AnnyKZ [126]
4 years ago
7

What is Index of refraction?

Physics
1 answer:
Sphinxa [80]4 years ago
3 0

Refractive Index is the measure of bending of light rays.

Explanation:

  • Index of refraction or Refractive Index is the measurement of the phenomenon exhibited by light rays - bending of light rays when passing from one medium to another.
  • It can also be measured as the ratio between speed of light in vacuum or empty space to that of the speed of light in a given medium.
  • Index of refraction is denoted by n and is calculated by the following formula

n = Speed of light in vacuum/Speed of light in the given medium = c/v

  • It can also be measured using the angle of incidence and angle of refraction.

n = sin i/sin r where i is the angle of incidence and r is the angle of refraction

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An element can be identified by _____.
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The answer is the number of protons in its atom

Explanation:

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Any help would be appreciated! :)​
ladessa [460]

Answer:

primary voltage/secondary voltage = number of turns on primary coil/number of turns on secondary coil

V1/V2= N1/N2

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Describe the relationship between the potential energy and kinetic energy of an object and the amount of Kinetic energy it can g
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The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

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It is the amount of work you have done in a certain amount of time. <span />
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