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1 year ago

a nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm. how many photons are in the pulse?

1 answer:

6 0

A nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm and has 1785 x 10¹⁹ photons in the pulse.

<h3>How many photons are in the pulse?</h3>

Energy of a single photon is

E=hcλ

E=6.626×10⁻³⁴ J s×3×108 m/s /340×10⁻⁹ m

E=6.31×10⁻¹⁹ J

Number of photons in the laser is

n=Total Energy/Energy per photon

n=10⁷×10⁻³J /5.90×10⁻¹⁹J/photon

n= 1785 x 10¹⁹ photons

To learn about photons, refer: brainly.com/question/20912241?referrer=searchResults

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Read 2 more answers

A 9.00-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wi

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Answer:

0.45 N

Explanation:

Let Recall that,

The power formula is:

P = E²/R

Let A = the magnetic field

Let L = length of wire = 9.00cm = 0.09 m

let R = resistance of wire = 0.320 Ω

let v = velocity of the wire = 4 m/s

Let E = across the wire voltage

Let P = the power of the wire = 4.3 W

To Solve for E:

The formula of E = √PR

The Voltage from a magnetic field is given as,

E = vAL

We therefore Use E = E

√PR = vAL

to solve for A,

A= √PR/vL

BA= √4.3(0.32)/(4)(.09) -=0.173

A = 0.173 wA/m²

Let F be the pulling force

Let I be the current in the wire

P = I²R

I = √P/R

F = IAL

F = √P/RAL

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3 years ago

The circuit in Fig. P4.23 utilizes three identical diodes having IS = 10−14 A. Find the value of the current I required to obtai

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Answer:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have: $I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

Explanation:

For this case we have the figure attached illustrating the problem

We know that the equation for the current in a diode id given by:

$I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}$

For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode $v_1 + v_2 + v_3= 2$ and each voltage is the same v for each diode, so then:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have:

$I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

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3 years ago