{"id":233412,"date":"2025-06-13T06:06:37","date_gmt":"2025-06-13T06:06:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233412"},"modified":"2025-06-13T06:06:39","modified_gmt":"2025-06-13T06:06:39","slug":"iron-fe-has-an-atomic-radius-of-126-pm","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/iron-fe-has-an-atomic-radius-of-126-pm\/","title":{"rendered":"Iron (Fe) has an atomic radius of 126 pm"},"content":{"rendered":"\n

Iron (Fe) has an atomic radius of 126 pm. What would the density of Fe be in g\/cm’ if it crystallized in the following crystal structures: Simple cubic (3 pts) Body centered cubic (3 pts)<\/p>\n\n\n\n

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

To find the density of iron (Fe)<\/strong> in g\/cm\u00b3<\/strong> under different crystal structures, we use the following formula:Density=Mass of atoms in unit cellVolume of unit cell\\text{Density} = \\frac{\\text{Mass of atoms in unit cell}}{\\text{Volume of unit cell}}Density=Volume of unit cellMass of atoms in unit cell\u200b<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Atomic radius of Fe = 126 pm = 1.26\u00d710\u221281.26 \\times 10^{-8}1.26\u00d710\u22128 cm<\/li>\n\n\n\n
  • Atomic mass of Fe = 55.85 g\/mol<\/li>\n\n\n\n
  • Avogadro\u2019s number NA=6.022\u00d71023N_A = 6.022 \\times 10^{23}NA\u200b=6.022\u00d71023 atoms\/mol<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    1. Simple Cubic (SC):<\/strong><\/h3>\n\n\n\n
      \n
    • Atoms per unit cell (SC)<\/strong> = 1<\/li>\n\n\n\n
    • Edge length<\/strong> a=2r=2\u00d71.26\u00d710\u22128\u00a0cm=2.52\u00d710\u22128\u00a0cma = 2r = 2 \\times 1.26 \\times 10^{-8} \\text{ cm} = 2.52 \\times 10^{-8} \\text{ cm}a=2r=2\u00d71.26\u00d710\u22128\u00a0cm=2.52\u00d710\u22128\u00a0cm<\/li>\n<\/ul>\n\n\n\n

      Step 1: Volume of unit cell:<\/h4>\n\n\n\n

      V=a3=(2.52\u00d710\u22128)3=1.60\u00d710\u221223 cm3V = a^3 = (2.52 \\times 10^{-8})^3 = 1.60 \\times 10^{-23} \\text{ cm}^3V=a3=(2.52\u00d710\u22128)3=1.60\u00d710\u221223 cm3<\/p>\n\n\n\n

      Step 2: Mass of one atom:<\/h4>\n\n\n\n

      Mass=55.856.022\u00d71023=9.28\u00d710\u221223 g\\text{Mass} = \\frac{55.85}{6.022 \\times 10^{23}} = 9.28 \\times 10^{-23} \\text{ g}Mass=6.022\u00d7102355.85\u200b=9.28\u00d710\u221223 g<\/p>\n\n\n\n

      Step 3: Density:<\/h4>\n\n\n\n

      Density=9.28\u00d710\u2212231.60\u00d710\u221223\u22485.8 g\/cm3\\text{Density} = \\frac{9.28 \\times 10^{-23}}{1.60 \\times 10^{-23}} \\approx 5.8 \\text{ g\/cm}^3Density=1.60\u00d710\u2212239.28\u00d710\u221223\u200b\u22485.8 g\/cm3<\/p>\n\n\n\n

      \n\n\n\n

      2. Body-Centered Cubic (BCC):<\/strong><\/h3>\n\n\n\n
        \n
      • Atoms per unit cell (BCC)<\/strong> = 2<\/li>\n\n\n\n
      • In BCC, edge length a=4r3=4\u00d71.26\u00d710\u221283=2.91\u00d710\u22128\u00a0cma = \\frac{4r}{\\sqrt{3}} = \\frac{4 \\times 1.26 \\times 10^{-8}}{\\sqrt{3}} = 2.91 \\times 10^{-8} \\text{ cm}a=3\u200b4r\u200b=3\u200b4\u00d71.26\u00d710\u22128\u200b=2.91\u00d710\u22128\u00a0cm<\/li>\n<\/ul>\n\n\n\n

        Step 1: Volume of unit cell:<\/h4>\n\n\n\n

        V=a3=(2.91\u00d710\u22128)3=2.46\u00d710\u221223 cm3V = a^3 = (2.91 \\times 10^{-8})^3 = 2.46 \\times 10^{-23} \\text{ cm}^3V=a3=(2.91\u00d710\u22128)3=2.46\u00d710\u221223 cm3<\/p>\n\n\n\n

        Step 2: Mass of 2 atoms:<\/h4>\n\n\n\n

        Mass=2\u00d755.856.022\u00d71023=1.856\u00d710\u221222 g\\text{Mass} = 2 \\times \\frac{55.85}{6.022 \\times 10^{23}} = 1.856 \\times 10^{-22} \\text{ g}Mass=2\u00d76.022\u00d7102355.85\u200b=1.856\u00d710\u221222 g<\/p>\n\n\n\n

        Step 3: Density:<\/h4>\n\n\n\n

        Density=1.856\u00d710\u2212222.46\u00d710\u221223\u22487.5 g\/cm3\\text{Density} = \\frac{1.856 \\times 10^{-22}}{2.46 \\times 10^{-23}} \\approx 7.5 \\text{ g\/cm}^3Density=2.46\u00d710\u2212231.856\u00d710\u221222\u200b\u22487.5 g\/cm3<\/p>\n\n\n\n

        \n\n\n\n

        Final Answers:<\/strong><\/h3>\n\n\n\n
          \n
        • Simple Cubic:<\/strong> \u2248 5.8 g\/cm\u00b3<\/strong><\/li>\n\n\n\n
        • Body-Centered Cubic:<\/strong> \u2248 7.5 g\/cm\u00b3<\/strong><\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Explanation (Textbook Style – 300 Words):<\/strong><\/h3>\n\n\n\n

          To determine the density of a substance based on its crystal structure, we need to calculate both the mass of atoms in a unit cell<\/strong> and the volume of that unit cell<\/strong>. The density formula, derived from mass\/volume, provides insight into how closely packed the atoms are within a specific crystal arrangement.<\/p>\n\n\n\n

          For iron (Fe)<\/strong>, which has an atomic radius of 126 picometers (pm)<\/strong>, we consider two possible crystal structures: simple cubic (SC)<\/strong> and body-centered cubic (BCC)<\/strong>.<\/p>\n\n\n\n

          In the simple cubic structure<\/strong>, only one atom occupies each unit cell. The edge of the cube is twice the atomic radius, giving an edge length of 2.52 \u00d7 10\u207b\u2078 cm<\/strong>. The volume of the cube is found by cubing the edge length. The mass of one atom is calculated using the atomic mass of iron and dividing by Avogadro\u2019s number. Dividing this atomic mass by the volume gives the density of about 5.8 g\/cm\u00b3<\/strong>.<\/p>\n\n\n\n

          In contrast, the body-centered cubic structure<\/strong> contains 2 atoms per unit cell<\/strong>. Its geometry is such that the cube\u2019s body diagonal equals 4 atomic radii<\/strong>, so the edge length is found using the relation a=4r3a = \\frac{4r}{\\sqrt{3}}a=3\u200b4r\u200b, resulting in a slightly larger cube. The volume is calculated by cubing this edge length. Since there are two atoms per cell, the total mass is doubled. Calculating the mass per volume gives a higher density, about 7.5 g\/cm\u00b3<\/strong>, reflecting the more efficient atomic packing of the BCC structure compared to the SC structure.<\/p>\n\n\n\n

          This analysis illustrates how the arrangement of atoms in space directly affects a metal’s density<\/strong>, even when atomic size and mass remain the same.<\/p>\n\n\n\n

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

          Iron (Fe) has an atomic radius of 126 pm. What would the density of Fe be in g\/cm’ if it crystallized in the following crystal structures: Simple cubic (3 pts) Body centered cubic (3 pts) The Correct Answer and Explanation is: To find the density of iron (Fe) in g\/cm\u00b3 under different crystal structures, we […]<\/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-233412","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233412","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=233412"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233412\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233412"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233412"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233412"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}