{"id":219820,"date":"2025-05-27T04:21:49","date_gmt":"2025-05-27T04:21:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219820"},"modified":"2025-05-27T04:21:51","modified_gmt":"2025-05-27T04:21:51","slug":"show-that-for-hexagonal-close-packed-structure-c-a-1-633-and-atomic-packing-factor-apf-0-74","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/27\/show-that-for-hexagonal-close-packed-structure-c-a-1-633-and-atomic-packing-factor-apf-0-74\/","title":{"rendered":"Show that, for Hexagonal Close-Packed Structure, c\/a = 1.633, and Atomic Packing Factor (APF) = 0.74"},"content":{"rendered":"\n<p>Show that, for Hexagonal Close-Packed Structure, c\/a = 1.633, and Atomic Packing Factor (APF) = 0.74<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-441.png\" alt=\"\" class=\"wp-image-219821\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To show that for a <strong>Hexagonal Close-Packed (HCP)<\/strong> structure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>c\/a = 1.633<\/strong>, and<\/li>\n\n\n\n<li><strong>Atomic Packing Factor (APF) = 0.74<\/strong>,<\/li>\n<\/ul>\n\n\n\n<p>we analyze the ideal geometry of an HCP unit cell.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>c\/a Ratio Derivation<\/strong><\/h3>\n\n\n\n<p>The HCP unit cell consists of 6 atoms and has an <strong>ABAB<\/strong> stacking sequence. The key is to consider the geometry formed by atoms in layers A and B.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Let the atomic radius be <strong>R<\/strong>.<\/li>\n\n\n\n<li>In the hexagonal base (A layer), atoms touch each other, forming an equilateral triangle:<br>So, side <strong>a = 2R<\/strong>.<\/li>\n\n\n\n<li>Now, consider the triangle formed by three atoms: one from the top A layer, one from the middle B layer, and one from the bottom A layer. The height between the two A layers is <strong>c<\/strong>, and we can analyze the tetrahedron formed by the 4 atoms: three in one plane and one directly above the center.<\/li>\n<\/ul>\n\n\n\n<p>Using geometry: c=223R=2R83=2R\u22c5263c = 2 \\sqrt{\\frac{2}{3}} R = 2R \\sqrt{\\frac{8}{3}} = 2R \\cdot \\frac{2\\sqrt{6}}{3}<\/p>\n\n\n\n<p>So, the <strong>c\/a ratio<\/strong> becomes: ca=2R\u22c52632R=263\u22481.633\\frac{c}{a} = \\frac{2R \\cdot \\frac{2\\sqrt{6}}{3}}{2R} = \\frac{2\\sqrt{6}}{3} \\approx 1.633<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Atomic Packing Factor (APF)<\/strong><\/h3>\n\n\n\n<p>APF is defined as the fraction of volume occupied by atoms in the unit cell: APF=Volume&nbsp;of&nbsp;atoms&nbsp;in&nbsp;unit&nbsp;cellVolume&nbsp;of&nbsp;unit&nbsp;cell\\text{APF} = \\frac{\\text{Volume of atoms in unit cell}}{\\text{Volume of unit cell}}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each unit cell has <strong>6 atoms<\/strong>.<\/li>\n\n\n\n<li>Volume of one atom: V=43\u03c0R3V = \\frac{4}{3} \\pi R^3<\/li>\n\n\n\n<li>Volume of 6 atoms: 6\u22c543\u03c0R3=8\u03c0R36 \\cdot \\frac{4}{3} \\pi R^3 = 8\\pi R^3<\/li>\n\n\n\n<li>Volume of unit cell: Vcell=Base\u00a0Area\u00d7c=(332a2)cV_{cell} = \\text{Base Area} \\times c = \\left(\\frac{3\\sqrt{3}}{2} a^2\\right)c Substituting a=2Ra = 2R, and c=1.633\u22c5ac = 1.633 \\cdot a: Vcell=332(2R)2(1.633\u22c52R)=332\u22c54R2\u22c53.266R=63\u22c53.266R3V_{cell} = \\frac{3\\sqrt{3}}{2} (2R)^2 (1.633 \\cdot 2R) = \\frac{3\\sqrt{3}}{2} \\cdot 4R^2 \\cdot 3.266R = 6\\sqrt{3} \\cdot 3.266 R^3<\/li>\n<\/ul>\n\n\n\n<p>Now calculate APF: APF=8\u03c0R363\u22c53.266R3\u22480.74\\text{APF} = \\frac{8\\pi R^3}{6\\sqrt{3} \\cdot 3.266 R^3} \\approx 0.74<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Result:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>c\/a = 1.633<\/strong><\/li>\n\n\n\n<li><strong>APF = 0.74<\/strong><\/li>\n<\/ul>\n\n\n\n<p>These values describe the ideal HCP structure, as shown in the slide.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner6-55.jpeg\" alt=\"\" class=\"wp-image-219822\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Show that, for Hexagonal Close-Packed Structure, c\/a = 1.633, and Atomic Packing Factor (APF) = 0.74 The Correct Answer and Explanation is: To show that for a Hexagonal Close-Packed (HCP) structure: we analyze the ideal geometry of an HCP unit cell. 1. c\/a Ratio Derivation The HCP unit cell consists of 6 atoms and has [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center 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