{"id":221755,"date":"2025-05-30T13:48:58","date_gmt":"2025-05-30T13:48:58","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221755"},"modified":"2025-05-30T13:49:01","modified_gmt":"2025-05-30T13:49:01","slug":"iron-has-a-bcc-crystal-structure-an-atomic-radius-of-0-124-nm-and-an-atomic-weight-of-55-85-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/30\/iron-has-a-bcc-crystal-structure-an-atomic-radius-of-0-124-nm-and-an-atomic-weight-of-55-85-g-mol\/","title":{"rendered":"Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g\/mol."},"content":{"rendered":"\n

Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g\/mol. Compute and compare its theoretical density with the experimental value. Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 g\/cm^3, and an atomic weight of 192.2 g\/mol. Calculate the radius of a vanadium atom, given vanadium has a BCC crystal structure, a density of 5.96 g\/cm^3, and an atomic weight of 50.9 g\/mol. Description: Iron has a BCC crystal structure, an atomic radius<\/p>\n\n\n\n

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

1. Theoretical Density of Iron (Fe)<\/strong><\/h3>\n\n\n\n

Given:<\/strong><\/p>\n\n\n\n

    \n
  • Structure: BCC (Body-Centered Cubic)<\/li>\n\n\n\n
  • Atomic radius (r) = 0.124 nm = 1.24 \u00d7 10\u207b\u2078 cm<\/li>\n\n\n\n
  • Atomic weight (A) = 55.85 g\/mol<\/li>\n\n\n\n
  • Avogadro\u2019s number (N\u2090) = 6.022 \u00d7 10\u00b2\u00b3 atoms\/mol<\/li>\n<\/ul>\n\n\n\n

    BCC Unit Cell Characteristics:<\/strong><\/p>\n\n\n\n

      \n
    • Number of atoms\/unit cell (n) = 2<\/li>\n\n\n\n
    • In BCC: a=4r3a = \\frac{4r}{\\sqrt{3}}<\/li>\n<\/ul>\n\n\n\n

      Step 1: Calculate unit cell volume VcV_c:<\/strong> a=4\u00d71.24\u00d710\u22128 cm3=2.864\u00d710\u22128 cma = \\frac{4 \\times 1.24 \\times 10^{-8} \\text{ cm}}{\\sqrt{3}} = 2.864 \\times 10^{-8} \\text{ cm} Vc=a3=(2.864\u00d710\u22128)3=2.35\u00d710\u221223 cm3V_c = a^3 = (2.864 \\times 10^{-8})^3 = 2.35 \\times 10^{-23} \\text{ cm}^3<\/p>\n\n\n\n

      Step 2: Theoretical Density \u03c1\\rho:<\/strong> \u03c1=n\u00d7AVc\u00d7Na=2\u00d755.852.35\u00d710\u221223\u00d76.022\u00d71023=7.87 g\/cm3\\rho = \\frac{n \\times A}{V_c \\times N_a} = \\frac{2 \\times 55.85}{2.35 \\times 10^{-23} \\times 6.022 \\times 10^{23}} = 7.87 \\text{ g\/cm}^3<\/p>\n\n\n\n

      \u2705 Result: Theoretical density of Fe = 7.87 g\/cm\u00b3<\/strong>, which matches well with the experimental value of 7.87 g\/cm\u00b3<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      2. Radius of Iridium (Ir)<\/strong><\/h3>\n\n\n\n

      Given:<\/strong><\/p>\n\n\n\n

        \n
      • Structure: FCC (Face-Centered Cubic)<\/li>\n\n\n\n
      • Density (\u03c1) = 22.4 g\/cm\u00b3<\/li>\n\n\n\n
      • Atomic weight (A) = 192.2 g\/mol<\/li>\n\n\n\n
      • n = 4 atoms\/unit cell (FCC)<\/li>\n<\/ul>\n\n\n\n

        Step 1: Volume of the unit cell:<\/strong> Vc=n\u22c5A\u03c1\u22c5Na=4\u22c5192.222.4\u22c56.022\u00d71023=5.714\u00d710\u221223 cm3V_c = \\frac{n \\cdot A}{\\rho \\cdot N_a} = \\frac{4 \\cdot 192.2}{22.4 \\cdot 6.022 \\times 10^{23}} = 5.714 \\times 10^{-23} \\text{ cm}^3<\/p>\n\n\n\n

        Step 2: Lattice parameter:<\/strong> a=Vc1\/3=(5.714\u00d710\u221223)1\/3=3.84\u00d710\u22128 cma = V_c^{1\/3} = (5.714 \\times 10^{-23})^{1\/3} = 3.84 \\times 10^{-8} \\text{ cm}<\/p>\n\n\n\n

        In FCC: a=22ra = 2\\sqrt{2}r \u2192 r=a22r = \\frac{a}{2\\sqrt{2}}:<\/strong> r=3.84\u00d710\u2212822=1.36\u00d710\u22128 cm=0.136 nmr = \\frac{3.84 \\times 10^{-8}}{2\\sqrt{2}} = 1.36 \\times 10^{-8} \\text{ cm} = 0.136 \\text{ nm}<\/p>\n\n\n\n

        \u2705 Radius of Ir atom = 0.136 nm<\/strong><\/p>\n\n\n\n

        \n\n\n\n

        3. Radius of Vanadium (V)<\/strong><\/h3>\n\n\n\n

        Given:<\/strong><\/p>\n\n\n\n

          \n
        • Structure: BCC<\/li>\n\n\n\n
        • \u03c1 = 5.96 g\/cm\u00b3<\/li>\n\n\n\n
        • A = 50.9 g\/mol<\/li>\n\n\n\n
        • n = 2 atoms\/unit cell<\/li>\n<\/ul>\n\n\n\n

          Step 1: Volume of the unit cell:<\/strong> Vc=2\u22c550.95.96\u22c56.022\u00d71023=2.83\u00d710\u221223 cm3V_c = \\frac{2 \\cdot 50.9}{5.96 \\cdot 6.022 \\times 10^{23}} = 2.83 \\times 10^{-23} \\text{ cm}^3<\/p>\n\n\n\n

          Step 2: Lattice parameter a=Vc1\/3a = V_c^{1\/3}:<\/strong> a=(2.83\u00d710\u221223)1\/3=3.05\u00d710\u22128 cma = (2.83 \\times 10^{-23})^{1\/3} = 3.05 \\times 10^{-8} \\text{ cm}<\/p>\n\n\n\n

          In BCC: a=4r3a = \\frac{4r}{\\sqrt{3}} \u2192 r=a34r = \\frac{a \\sqrt{3}}{4}:<\/strong> r=3.05\u00d710\u22128\u22c534=1.32\u00d710\u22128 cm=0.132 nmr = \\frac{3.05 \\times 10^{-8} \\cdot \\sqrt{3}}{4} = 1.32 \\times 10^{-8} \\text{ cm} = 0.132 \\text{ nm}<\/p>\n\n\n\n

          \u2705 Radius of V atom = 0.132 nm<\/strong><\/p>\n\n\n\n

          \n\n\n\n

          Conclusion (Summary Table)<\/strong><\/h3>\n\n\n\n
          Element<\/th>Structure<\/th>Atomic Radius (nm)<\/th>Theoretical Density (g\/cm\u00b3)<\/th><\/tr><\/thead>
          Iron (Fe)<\/td>BCC<\/td>0.124<\/td>7.87<\/td><\/tr>
          Iridium (Ir)<\/td>FCC<\/td>0.136<\/td>22.4 (given)<\/td><\/tr>
          Vanadium (V)<\/td>BCC<\/td>0.132<\/td>5.96 (given)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          These calculations show how atomic structure, radius, and density interrelate in crystalline materials, helping us understand their physical properties.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g\/mol. Compute and compare its theoretical density with the experimental value. Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 g\/cm^3, and an atomic weight of 192.2 […]<\/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-221755","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221755","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=221755"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221755\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221755"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221755"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221755"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}