Exercise 1.1 E=1.602×10−16J,p=1.708×10−23kg⋅m/s,λ=3.879×10−11mE=1.602×10−16J,p=1.708×10−23kg⋅m/s,λ=3.879×10−11mE = 1.602 {\textbackslash}times 1{0\}{\textasciicircum}{-16\}{\textbackslash},{\textbackslash}mathrm{J},p = 1.708 {\textbackslash}times 1{0\}{\textasciicircum}{-23\}{\textbackslash},{\textbackslash}mathrm{kg} {\textbackslash}cdot {\textbackslash}mathrm{m/s},{\textbackslash}lambda = 3.879 {\textbackslash}times 1{0\}{\textasciicircum}{-11\}{\textbackslash},{\textbackslash}mathrm{m}
CITATION STYLE
Waseda, Y., Matsubara, E., & Shinoda, K. (2011). Solutions to Supplementary Problems. In X-Ray Diffraction Crystallography (pp. 273–288). Springer Berlin Heidelberg. https://doi.org/10.1007/978-3-642-16635-8_8
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