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KIM [24]
2 years ago
6

A grating has 470 lines/mm. how many orders of the visible wavelength 538 nm can it produce in addition to the m = 0 order?

Physics
1 answer:
Natali [406]2 years ago
4 0

Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.

The ruling separation is d=1/(470mm-1)

= 2.1 \times 10 {}^{ - 3}mm

Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.

Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.

we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.

That is,find the greater integer value of m for which mλ<d.

since,d/λ

= 538 \times 10 {}^{ - 9} m/2.1 \times 10 {}^{ - 6}

There are three complete orders on each side of the m=0 order.

The second and third orders overlap.

learn more about diffraction from here: brainly.com/question/28168352

#SPJ4

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Explanation:

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As we go up in atomic number, the energy released per nuclei goes down until it hits a minimum which is for atomic number 26 (iron) and fusion is not possible.

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3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

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at t = 0.01

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at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

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now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
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maksim [4K]

Answer:

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At the closure temperature, that rate of diffusion will be zero as the isotopes will be no longer free to move out of crystal lattice. So, this is termed as closure or blocking temperature. As the isotopes loose their ability to move, their concentration will remain fixed in their position leading to measurement of radiation dating.

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