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lina2011 [118]
3 years ago
15

In LASIK surgery, a laser is used to reshape the cornea of the eye to improve vision. The laser produces extremely short pulses

of light, each containing 1.0 mJ of energy. Part A In each pulse there are 9.7×1014 photons. What is the wavelength of the laser?
Physics
1 answer:
qwelly [4]3 years ago
8 0

To solve this problem it is necessary to apply the related concepts laser pulse energy.

By definition the energy of a laser pulse in terms of number of photons is

E = N \frac{hc}{\lambda}

Where,

\lambda = Wavelength

N = Number of photons in each laser pulse

h = Planck constant

c = Speed of light

We need to find the wavelength, then re-arrange the equation we have

\lambda = \frac{Nhc}{E}

Converting the unit the energy from J to eV, we have

E= 1*10^{-3}J(\frac{1eV}{1.6*10^{-19}J})

E = 6.25*10^{15}eV

Replacing,

\lambda = \frac{Nhc}{E}

\lambda = \frac{(9.7*10^{14})(4.136*10^{-15})(3*10^8)}{6.25*10^{15}}

\lambda = 192.4nm

Therefore the wavelength of the laser is 182.4nm

<em>Note: The Planck constant used is in units of eV.</em>

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Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
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