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3 years ago

A 193nm-wavelength UV laser for eye surgery emits a 0.500mJ pulse. (a) How many photons does the light pulse contain?

Physics

2 answers:

Law Incorporation [45]3 years ago

6 0

Answer:

The number of photons is 4.8x10^14

Explanation:

The frequency of wave is equal to:

$f=\frac{c}{l}$

where c is the speed of light, l is the wavelength of wave. Replacing values:

$f=\frac{3x10^{8} }{193x10^{9} } =1.5x10^{15} Hz$

The energy of the proton is:

$E=hf$

where h is the Planck´s constant. Replacing

$E=6.626x10^{-34}*1.5x10^{5}=1.03x10^{-18} J$

The number of photons is:

$n=\frac{E1}{E}$

where E1 is the energy of photon. Replacing:

$n=\frac{0.5x10^{-3} }{1.03x10^{-18} }=4.8x10^{14}$

Marrrta [24]3 years ago

4 0

Using the equation E = hc/λ we can find out how much energy a single photon of wavelength 193 nm has. E = Planck Constant * Speed of Light/193 nm

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Answer:

The peak emf of the generator is 40.94 V.

Explanation:

Given that,

Number of turns in primary coil= 11

Number of turns in secondary coil= 18

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Using relation of number of turns and emf

$\dfrac{N_{1}}{N_{2}}=\dfrac{E_{1}}{E_{2}}$

$E_{1}=\dfrac{N_{1}}{N_{2}}\times E_{2}$

Where, N₁ = Number of turns in primary coil

N₂ = Number of turns in secondary coil

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Put the value into the formula

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2 years ago

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Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

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$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

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Answer:

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Explanation:

Below is an attachment containing the solution.