If an x-ray with a of 164 pm is at an according to the Bragg equation (nlambda -

Question

If an x-ray with a of 164 pm is at an according to the Bragg equation (nlambda - 2 d sintheta) what is the distance between the crystal that give rise to this diffraction? For this problem n = 1. Use some data to estimate the Si-Si stretching wavenumber tor the compound Si_2^2- and draw the molecular orbital diagram to explain the values For the species draw the structure and indicate and draw (if there is) all possible isomers Using the ligand ortho-phenylenediamine indicate and draw the structures of the possible complexes synthesized from CoCl_3-6H_2O Indicate and explain the trend in oxidation states between Halides of Cu, Ag, Au Which of the following have preference to form oxides and which to form and explain why? V(V) and Co(II) For the following complexes, identify the one that has the larger LFSE (express m terms of C_g, t_2g or e, t_2 and pairing energy E) [Ru(H_2o)_6] [Ru(pyndine) [Ru(bipyridinc)_3]^2+ Draw the molecular orbital diagram (showing s, p and d orbitals, sigma, pi and delt orbitals for the complex [Mn(pyridine)_6]^2+ Draw the structure of KTaO, and indicate if there is room (and where it is) to host a Horn or ion Explain vapor-pha se synthesis of oxides nanoparticles method and give one example

Explanation / Answer

Given data:

= 164 pm = 164 x 10-12 m = 1.64 x 10-10 m.

Angle of diffraction 2 = 22.8o

So, = 11.4o

n = 1

Distance between crystal layers = d = ?

Bragg’s equation is given as,

n = 2 d sin

1 x (1.64 x 10-10 ) = 2 d sin(11.4o)

1.64 x 10-10 = 2 d x 0.198

1.64 x 10-10 = d x 0.396

d = 1.64 x 10-10/0.396

d = 4.414 x 10-10 m

d = 4.414 Å

d = 414.4 pm

Distance between crystal layers is 4.414 Å or 414.4 pm

======================XXXXXXXXXXXXXXX===========================

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---