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Using d sin π = nπ, d=98.2pm, n=1, π=17.5ΒΊ

98.2*sin(17.5ΒΊ) = 1*π

π=29.53pm

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Q: A crystal has an interplanar spacing of 98.2 pm. Using first order diffraction, what is the X-ray wavelength required to produce a reflection with an angle of 17.5 degrees?

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Simply using the relation wavelength = velocity / frequency So required wavelength = 25/5 = 5 m

Wavelength also increases proportionally. Reasoning: Relevant equation- v= f x wavelength therefore speed is directly proportional to wavelength, and so as speed increases, the wavelength would increase proportionally. hope this helps if more information is required, email me @ physicsisland@hotmail.com

it is required to improve the efficiency of production.

Joining Period Required?

The word "expected" is not the same as "required". Something that is "expected" is something that is assumed will occur. Something that is "required" is something that is essential.

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To get the required wavelength of light

Simply using the relation wavelength = velocity / frequency So required wavelength = 25/5 = 5 m

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Wavelength also increases proportionally. Reasoning: Relevant equation- v= f x wavelength therefore speed is directly proportional to wavelength, and so as speed increases, the wavelength would increase proportionally. hope this helps if more information is required, email me @ physicsisland@hotmail.com

reflection just means the mirror image. in the coordinate system just reflect the graph across the given axis that y ou are required.

Speed = frequency x wavelength So required speed = 0.5 * 1 = 0.5 m/s Problem is that the wavelength is not given, but I have taken as 1 m for granted. Hence the answer

The formula related to frequency and wavelength is Wavelength = 300000000 / Frequency (f) Wavelength = 300000000 / 30000000000 Wavelength = 1/100 Wavelength = 0.01 meter OR Wavelength = 10 milimeter

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Simply using the formula wavelength = velocity of the wave / frequency we can get the required. So we need velocity of sound wave. Usually we take it to be 330 m/s at room temperature. Now using 330 and 21 we can get wavelength in metre