{"id":45854,"date":"2025-07-01T13:21:18","date_gmt":"2025-07-01T13:21:18","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=45854"},"modified":"2025-07-01T13:21:19","modified_gmt":"2025-07-01T13:21:19","slug":"the-unified-atomic-mass-unit-is-defined-to-be-1-u-1-6605x10%e2%88%9227-kg","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-unified-atomic-mass-unit-is-defined-to-be-1-u-1-6605x10%e2%88%9227-kg\/","title":{"rendered":"The unified atomic mass unit is defined to be 1 u =\u00a01.6605\u00d710\u221227\u00a0kg."},"content":{"rendered":"\n
The unified atomic mass unit is defined to be 1 u = 1.6605\u00d710\u221227 kg.<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Based on the provided calculation, the correct value for the final blank box is 9.315e8<\/strong>.<\/p>\n\n\n\n
Explanation<\/h3>\n\n\n\n
The problem requires verifying the energy equivalence of one unified atomic mass unit (u), which is stated to be 931.5 Mega-electron-volts (MeV). This verification process is based on Albert Einstein’s famous\u207b\u00b9\u2079 J\/eV)
Energy (eV) \u2248 931,500,000 eV, which is written in scientific notation as 9.315 x 10\u2078 eV.<\/p>\n\n\n\n
This value, 9.315 x 10\u2078 eV, is the correct entry for the box that was marked as incorrect in the image.<\/p>\n\n\n\n
The final step is to convert the energy from electron-volts to mega-electron-volts. The prefix “mega” means one million, so 1 MeV = 10\u2076 eV. We divide the energy in eV by 10\u2076:<\/p>\n\n\n\n
Energy (MeV) = (9.315 x 10\u2078 eV) \/ (10\u2076 eV\/MeV) = 931.5 MeV.<\/p>\n\n\n\n
This calculation successfully verifies that 1 u of mass is equivalent to 931.5 MeV of energy. mass-energy equivalence principle, described by the equation E = mc\u00b2.<\/p>\n\n\n\n
Step 1: Calculate the Energy in Joules<\/strong><\/p>\n\n\n\n
The first step is to calculate the energy (E) in the standard SI unit, Joules (J). We use the given mass m = 1.6605 x 10\u207b\u00b2\u2077 kg for 1 u. The problem specifies using a value for the speed of light (c) with at least four-digit precision. The calculation in the image uses c = 2.998 x 10\u2078 m\/s.<\/p>\n\n\n\n
Plugging these values into the formula:
E = (1.6605 x 10\u207b\u00b2\u2077 kg) * (2.998 x 10\u2078 m\/s)\u00b2
E \u2248 1.4925 x 10\u207b\u00b9\u2070 J<\/p>\n\n\n\n
This result matches the first calculated value in the image, confirming the initial step is correct.<\/p>\n\n\n\n
Step 2: Convert Joules to Electron-Volts (eV)<\/strong><\/p>\n\n\n\n
For energies at the atomic and subatomic levels, the Joule is an inconveniently large unit. A more common unit is the electron-volt (eV). The conversion factor between Joules and electron-volts is the elementary charge, |q_e|. The problem uses a high-precision value for this conversion: 1 eV = 1.60218 x 10\u207b\u00b9\u2079 J.<\/p>\n\n\n\n
To convert the energy from Joules to electron-volts, we divide the energy in Joules by this conversion factor:
E (in eV) = (1.4925 x 10\u207b\u00b9\u2070 J) \/ (1.60218 x 10\u207b\u00b9\u2079 J\/eV)
E \u2248 931,550,000 eV or 9.3155 x 10\u2078 eV<\/p>\n\n\n\n
This result, 9.3155 x 10\u2078, is the correct value for the final blank box. The image shows that an incorrect value of 9.315e6 was entered, which is off by a factor of 100. The correct exponent is 8.<\/p>\n\n\n\n
Step 3: Convert Electron-Volts to Mega-electron-volts (MeV)<\/strong><\/p>\n\n\n\n
The final step is to express this energy in Mega-electron-volts. The prefix “Mega” (M) represents a factor of one million, or 10\u2076. To convert from eV to MeV, we divide by 10\u2076:
E (in MeV) = (9.3155 x 10\u2078 eV) \/ (10\u2076 eV\/MeV)
E \u2248 931.55 MeV<\/p>\n\n\n\n
Rounding this to one decimal place gives 931.5 MeV, which successfully verifies the initial statement.<\/p>\n\n\n\n
\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
The unified atomic mass unit is defined to be 1 u = 1.6605\u00d710\u221227 kg. The Correct Answer and Explanation is: Based on the provided calculation, the correct value for the final blank box is 9.315e8. Explanation The problem requires verifying the energy equivalence of one unified atomic mass unit (u), which is stated to be 931.5 Mega-electron-volts (MeV). […]<\/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-45854","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45854","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=45854"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45854\/revisions"}],"predecessor-version":[{"id":45858,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45854\/revisions\/45858"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=45854"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=45854"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=45854"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}