{"id":211120,"date":"2025-05-03T08:37:44","date_gmt":"2025-05-03T08:37:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=211120"},"modified":"2025-05-03T08:37:47","modified_gmt":"2025-05-03T08:37:47","slug":"calculate-the-planar-atomic-density-in-atoms-per-square-millimeter-for-the-following-crystal-planes-in-bcc-chromium-which-has-a-lattice-constant-of-0-28846-nm","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/03\/calculate-the-planar-atomic-density-in-atoms-per-square-millimeter-for-the-following-crystal-planes-in-bcc-chromium-which-has-a-lattice-constant-of-0-28846-nm\/","title":{"rendered":"Calculate the planar atomic density in atoms per square millimeter for the following crystal planes in BCC chromium, which has a lattice constant of 0.28846 nm."},"content":{"rendered":"\n

Calculate the planar atomic density in atoms per square millimeter for the following crystal planes in BCC chromium, which has a lattice constant of 0.28846 nm. Compare the values and draw a conclusion. (a) (100), (b) (110), (c) (111).<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To calculate planar atomic density (PAD)<\/strong> for different crystal planes in body-centered cubic (BCC) chromium<\/strong>, we use the formula: PAD=Number of atoms centered on the planeArea of the plane\\text{PAD} = \\frac{\\text{Number of atoms centered on the plane}}{\\text{Area of the plane}}PAD=Area of the planeNumber of atoms centered on the plane\u200b<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Crystal structure: BCC<\/strong><\/li>\n\n\n\n
  • Lattice constant, a=0.28846\u2009nm=0.28846\u00d710\u22126\u2009mma = 0.28846 \\, \\text{nm} = 0.28846 \\times 10^{-6} \\, \\text{mm}a=0.28846nm=0.28846\u00d710\u22126mm<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    (a) Plane (100)<\/strong><\/h2>\n\n\n\n
      \n
    • Atoms per (100) plane: Each corner contributes 1\/4 to the plane and there are 4 corners in the square face. So: Atoms\u00a0on\u00a0(100)=4\u00d714=1\\text{Atoms on (100)} = 4 \\times \\frac{1}{4} = 1Atoms\u00a0on\u00a0(100)=4\u00d741\u200b=1<\/li>\n\n\n\n
    • Area of (100) plane<\/strong>: Square with side aaa: A=a2=(0.28846\u00d710\u22126)2=8.322\u00d710\u221214\u2009mm2A = a^2 = (0.28846 \\times 10^{-6})^2 = 8.322 \\times 10^{-14} \\, \\text{mm}^2A=a2=(0.28846\u00d710\u22126)2=8.322\u00d710\u221214mm2<\/li>\n\n\n\n
    • PAD (100)<\/strong>: PAD(100)=18.322\u00d710\u221214\u22481.201\u00d71013\u2009atoms\/mm2\\text{PAD}_{(100)} = \\frac{1}{8.322 \\times 10^{-14}} \\approx 1.201 \\times 10^{13} \\, \\text{atoms\/mm}^2PAD(100)\u200b=8.322\u00d710\u2212141\u200b\u22481.201\u00d71013atoms\/mm2<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      (b) Plane (110)<\/strong><\/h2>\n\n\n\n
        \n
      • Atoms on (110): 4 corners \u00d7 1\/4 + 1 body atom in plane = 1+1=21 + 1 = 21+1=2<\/li>\n\n\n\n
      • Area of (110): Rectangle with sides aaa and a2a\\sqrt{2}a2\u200b: A=a\u00d7a2=a22=8.322\u00d710\u221214\u00d72=1.177\u00d710\u221213\u2009mm2A = a \\times a\\sqrt{2} = a^2 \\sqrt{2} = 8.322 \\times 10^{-14} \\times \\sqrt{2} = 1.177 \\times 10^{-13} \\, \\text{mm}^2A=a\u00d7a2\u200b=a22\u200b=8.322\u00d710\u221214\u00d72\u200b=1.177\u00d710\u221213mm2<\/li>\n\n\n\n
      • PAD (110)<\/strong>: PAD(110)=21.177\u00d710\u221213\u22481.699\u00d71013\u2009atoms\/mm2\\text{PAD}_{(110)} = \\frac{2}{1.177 \\times 10^{-13}} \\approx 1.699 \\times 10^{13} \\, \\text{atoms\/mm}^2PAD(110)\u200b=1.177\u00d710\u2212132\u200b\u22481.699\u00d71013atoms\/mm2<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        (c) Plane (111)<\/strong><\/h2>\n\n\n\n
          \n
        • Atoms on (111): 3 corners \u00d7 1\/6 + 1 body-centered atom = 0.5+1=1.50.5 + 1 = 1.50.5+1=1.5<\/li>\n\n\n\n
        • Area of (111): Equilateral triangle with side a2a\\sqrt{2}a2\u200b A=34(a2)2=34\u00d72a2=32\u00d7a2A = \\frac{\\sqrt{3}}{4}(a\\sqrt{2})^2 = \\frac{\\sqrt{3}}{4} \\times 2a^2 = \\frac{\\sqrt{3}}{2} \\times a^2A=43\u200b\u200b(a2\u200b)2=43\u200b\u200b\u00d72a2=23\u200b\u200b\u00d7a2 A=32\u00d78.322\u00d710\u221214=7.208\u00d710\u221214\u2009mm2A = \\frac{\\sqrt{3}}{2} \\times 8.322 \\times 10^{-14} = 7.208 \\times 10^{-14} \\, \\text{mm}^2A=23\u200b\u200b\u00d78.322\u00d710\u221214=7.208\u00d710\u221214mm2<\/li>\n\n\n\n
        • PAD (111)<\/strong>: PAD(111)=1.57.208\u00d710\u221214\u22482.080\u00d71013\u2009atoms\/mm2\\text{PAD}_{(111)} = \\frac{1.5}{7.208 \\times 10^{-14}} \\approx 2.080 \\times 10^{13} \\, \\text{atoms\/mm}^2PAD(111)\u200b=7.208\u00d710\u2212141.5\u200b\u22482.080\u00d71013atoms\/mm2<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Summary:<\/strong><\/h2>\n\n\n\n
          Plane<\/th>PAD (atoms\/mm\u00b2)<\/th><\/tr><\/thead>
          (100)<\/td>1.201\u00d710131.201 \\times 10^{13}1.201\u00d71013<\/td><\/tr>
          (110)<\/td>1.699\u00d710131.699 \\times 10^{13}1.699\u00d71013<\/td><\/tr>
          (111)<\/td>2.080\u00d710132.080 \\times 10^{13}2.080\u00d71013<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
          \n\n\n\n

          Conclusion:<\/strong><\/h3>\n\n\n\n

          The (111) plane<\/strong> has the highest planar atomic density<\/strong>, followed by (110)<\/strong>, and then (100)<\/strong>. This reflects how closely packed atoms are on each plane. In BCC structures, although (111) is not the densest direction (unlike FCC), it still contains more atoms per unit area than the other planes. This affects properties like slip system activation and surface energy\u2014important in deformation and materials engineering.<\/p>\n","protected":false},"excerpt":{"rendered":"

          Calculate the planar atomic density in atoms per square millimeter for the following crystal planes in BCC chromium, which has a lattice constant of 0.28846 nm. Compare the values and draw a conclusion. (a) (100), (b) (110), (c) (111). The Correct Answer and Explanation is: To calculate planar atomic density (PAD) for different crystal planes […]<\/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 center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-211120","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211120","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=211120"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211120\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211120"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211120"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211120"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}