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Andrei [34K]
2 years ago
14

The structure of the NaCl crystal forms reflecting planes 0.541 nm apart. What is the smallest angle, measured from these planes

, at which constructive interference of an X-ray beam reflecting off the two planes is observed?
Physics
1 answer:
elena55 [62]2 years ago
4 0

Answer:

\theta=4.5^{\circ}

Explanation:

Given that,

The distance between reflecting planes is 0.541 nm, d = 0.541 nm

Let the wavelength is 0.085 nm

We need to find the smallest angle, measured from these planes, at which constructive interference. Using Bragg's equation to find it as follows :

m\lambda=2d\sin\theta

For smallest angle, m = 1

\theta=\sin^{-1}(\dfrac{m\lambda}{2d})\\\\\theta=\sin^{-1}(\dfrac{1\times 0.085\ nm }{2\times 0.541\ nm })\\\\\theta=4.5^{\circ}

So, the smallest angle is 4.5 degrees.

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Explanation:

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3 years ago
An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, excit
worty [1.4K]

Answer:

Explanation:

An electron accelerates through a 12.5 V potential difference, starting from rest, so it will acquire kinetic energy of 12.5 eV .

In hydrogen atom energy of n th orbit in terms of eV is given as follows

En = -13.6 / n² eV

Total energy of 1 st orbit  E₁ = - 13.6 eV

Total energy of 2 st orbit E₂ = - 13.6 eV / 2² = - 3.4 eV

Total energy of 3 st orbit E₃ = - 13.6 eV / 3² = - 1.5  eV

Total energy of 4 st orbit E₄ = - 13.6 eV / 4² = - 0.85 eV

E₄ - E₁ = 13.6 - 0.85 = 12.75 eV

E₃ - E₁ = 13.6 - 1.5 = 12.10 eV

E₂ - E₁ = 13.6 - 3.4 = 10.2 eV .

The electron has energy of 12,5 eV so it can excite electron from E₁ to E₃ . .

Jump possible = E₃ to E₂ , E₂ to E₁ and E₃ to E₁

Energy of E₃ to E₂ = 3.4 - 1.5 eV = 1.9 eV

wavelength = 1237 / 1.9 nm = 651 nm

E₃ - E₁ = 13.6 - 1.5 = 12.10 eV

wavelength  = 1237 / 12.10  nm = 102.23 nm

E₂ - E₁ = 13.6 - 3.4 = 10.2 eV

wavelength  = 1237 / 10.2  nm = 121.27  nm

wavelength of photon possible are 651 nm , 121.27  nm , 102.23 nm .

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2 years ago
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