{"id":40993,"date":"2025-06-28T03:21:39","date_gmt":"2025-06-28T03:21:39","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=40993"},"modified":"2025-06-28T03:21:40","modified_gmt":"2025-06-28T03:21:40","slug":"silicon-has-a-diamond-diamond-crystal-structure-and-has-a-lattice-constant-a-0-543-nm","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/silicon-has-a-diamond-diamond-crystal-structure-and-has-a-lattice-constant-a-0-543-nm\/","title":{"rendered":"Silicon has a diamond (Diamond) crystal structure and has a lattice constant a = 0.543 nm"},"content":{"rendered":"\n<p>Silicon has a diamond (Diamond) crystal structure and has a lattice constant a = 0.543 nm. Atomic mass of silicon Considering that silicon is 28.08 g \/ mol, (a) the atomic density per unit volume (b) of silicon density g \/ cm3 Find in terms of.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer<\/strong><\/p>\n\n\n\n<p><strong>Given Data:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Silicon has a diamond cubic structure<\/li>\n\n\n\n<li>Lattice constant, <strong>a = 0.543 nm = 0.543 \u00d7 10\u207b\u2077 cm<\/strong><\/li>\n\n\n\n<li>Atomic mass of silicon = <strong>28.08 g\/mol<\/strong><\/li>\n\n\n\n<li>Avogadro\u2019s number = <strong>6.022 \u00d7 10\u00b2\u00b3 atoms\/mol<\/strong><\/li>\n\n\n\n<li>Number of atoms per unit cell in diamond cubic structure = <strong>8 atoms<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(a) Atomic Density per Unit Volume<\/strong><\/h3>\n\n\n\n<p>Atomic density = (Number of atoms per unit cell) \u00f7 (Volume of unit cell)<\/p>\n\n\n\n<p><strong>Volume of unit cell = a\u00b3<\/strong>a=0.543\u00d710\u22127\u2009cma = 0.543 \\times 10^{-7} \\, \\text{cm} a=0.543\u00d710\u22127cmVolume=a3=(0.543\u00d710\u22127)3=1.602\u00d710\u221222\u2009cm3\\text{Volume} = a^3 = (0.543 \\times 10^{-7})^3 = 1.602 \\times 10^{-22} \\, \\text{cm}^3 Volume=a3=(0.543\u00d710\u22127)3=1.602\u00d710\u221222cm3<\/p>\n\n\n\n<p>Total atoms per unit cell = 8<\/p>\n\n\n\n<p>Atomic density:Atomic&nbsp;Density=81.602\u00d710\u221222=4.995\u00d71022\u2009atoms\/cm3\\text{Atomic Density} = \\frac{8}{1.602 \\times 10^{-22}} = 4.995 \\times 10^{22} \\, \\text{atoms\/cm}^3 Atomic&nbsp;Density=1.602\u00d710\u2212228\u200b=4.995\u00d71022atoms\/cm3<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(b) Density of Silicon in g\/cm\u00b3<\/strong><\/h3>\n\n\n\n<p>Density formula:Density=Mass&nbsp;of&nbsp;atoms&nbsp;in&nbsp;one&nbsp;unit&nbsp;cellVolume&nbsp;of&nbsp;unit&nbsp;cell\\text{Density} = \\frac{\\text{Mass of atoms in one unit cell}}{\\text{Volume of unit cell}} Density=Volume&nbsp;of&nbsp;unit&nbsp;cellMass&nbsp;of&nbsp;atoms&nbsp;in&nbsp;one&nbsp;unit&nbsp;cell\u200b<\/p>\n\n\n\n<p><strong>Mass of one silicon atom<\/strong>:Mass&nbsp;per&nbsp;atom=28.08\u2009g\/mol6.022\u00d71023\u2009atoms\/mol=4.662\u00d710\u221223\u2009g\\text{Mass per atom} = \\frac{28.08 \\, \\text{g\/mol}}{6.022 \\times 10^{23} \\, \\text{atoms\/mol}} = 4.662 \\times 10^{-23} \\, \\text{g} Mass&nbsp;per&nbsp;atom=6.022\u00d71023atoms\/mol28.08g\/mol\u200b=4.662\u00d710\u221223g<\/p>\n\n\n\n<p>Total mass in one unit cell:Total&nbsp;mass=8\u00d74.662\u00d710\u221223=3.7296\u00d710\u221222\u2009g\\text{Total mass} = 8 \\times 4.662 \\times 10^{-23} = 3.7296 \\times 10^{-22} \\, \\text{g} Total&nbsp;mass=8\u00d74.662\u00d710\u221223=3.7296\u00d710\u221222g<\/p>\n\n\n\n<p>Density:Density=3.7296\u00d710\u2212221.602\u00d710\u221222=2.33\u2009g\/cm3\\text{Density} = \\frac{3.7296 \\times 10^{-22}}{1.602 \\times 10^{-22}} = 2.33 \\, \\text{g\/cm}^3 Density=1.602\u00d710\u2212223.7296\u00d710\u221222\u200b=2.33g\/cm3<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answers<\/strong><\/h3>\n\n\n\n<p><strong>(a)<\/strong> Atomic density = <strong>4.995 \u00d7 10\u00b2\u00b2 atoms\/cm\u00b3<\/strong><br><strong>(b)<\/strong> Density of silicon = <strong>2.33 g\/cm\u00b3<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong> Explanation<\/strong><\/h3>\n\n\n\n<p>Silicon crystallizes in a diamond cubic structure which is a common structure for group IV elements like carbon, silicon, and germanium. In this arrangement, each silicon atom is covalently bonded to four neighboring silicon atoms forming a tetrahedral geometry. The unit cell of this structure contains eight atoms.<\/p>\n\n\n\n<p>To calculate atomic density, we first need to determine the volume of the cubic unit cell. The lattice constant <strong>a<\/strong> is the edge length of the cube, given as 0.543 nanometers which converts to <strong>0.543 \u00d7 10\u207b\u2077 centimeters<\/strong>. The volume of the cube is obtained by raising this length to the power of three.<\/p>\n\n\n\n<p>Once the volume is known, the atomic density is found by dividing the number of atoms per unit cell by the unit cell&#8217;s volume. This gives the number of atoms present per cubic centimeter of silicon crystal.<\/p>\n\n\n\n<p>To calculate the density of silicon in grams per cubic centimeter, the mass of one silicon atom is required. The atomic mass is given as <strong>28.08 grams per mole<\/strong>, and dividing this by Avogadro\u2019s number provides the mass of one silicon atom. Multiplying the mass of one atom by the total number of atoms per unit cell gives the mass of the unit cell. Finally, dividing this mass by the unit cell&#8217;s volume provides the density.<\/p>\n\n\n\n<p>The final calculated density of silicon matches the known experimental value of approximately <strong>2.33 g\/cm\u00b3<\/strong>, which confirms the consistency of this method. This information is essential in semiconductor physics, materials science, and engineering applications where silicon is widely used.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-264.jpeg\" alt=\"\" class=\"wp-image-40994\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-264.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-264-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-264-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-264-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Silicon has a diamond (Diamond) crystal structure and has a lattice constant a = 0.543 nm. Atomic mass of silicon Considering that silicon is 28.08 g \/ mol, (a) the atomic density per unit volume (b) of silicon density g \/ cm3 Find in terms of. The Correct Answer and Explanation is: Correct Answer Given [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-40993","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40993","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=40993"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40993\/revisions"}],"predecessor-version":[{"id":40995,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40993\/revisions\/40995"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=40993"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=40993"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=40993"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}