{"id":149269,"date":"2024-03-04T17:46:52","date_gmt":"2024-03-04T12:16:52","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=149269"},"modified":"2024-03-04T17:46:52","modified_gmt":"2024-03-04T12:16:52","slug":"the-fission-properties-of-pu239-are-very-similar-to-those-of-92u235-how","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/quiz\/the-fission-properties-of-pu239-are-very-similar-to-those-of-92u235-how","title":{"rendered":"The fission properties of Pu239 are very similar to those of 92U235. How…"},"content":{"rendered":"

The fission properties of Pu239 are very similar to those of 92U235. How much energy (in MeV), is released if all the atoms in 1 g of pure 94Pu239 undergo fission? The average energy released per fission is 180 MeV.<\/h2>\n

Ans.<\/h3>\n

To find the energy released when all the atoms in 1 g of pure 94\ufffd\ufffd239<\/span>94<\/span><\/span><\/span><\/span><\/span><\/span><\/span>P<\/span>u<\/span>239<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> undergo fission, we need to calculate the number of atoms in 1 g of 94\ufffd\ufffd239<\/span>94<\/span><\/span><\/span><\/span><\/span><\/span><\/span>P<\/span>u<\/span>239<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> and then multiply it by the average energy released per fission.<\/h3>\n

First, we need to find the number of moles of 94\ufffd\ufffd239<\/span>94<\/span><\/span><\/span><\/span><\/span><\/span><\/span>P<\/span>u<\/span>239<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> in 1 g using its molar mass.<\/h3>\n

The molar mass of 94\ufffd\ufffd239<\/span>94<\/span><\/span><\/span><\/span><\/span><\/span><\/span>P<\/span>u<\/span>239<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is approximately 239\u2009g\/mol<\/span>239<\/span>g\/mol<\/span><\/span><\/span><\/span><\/span><\/span>.<\/h3>\n

So, the number of moles of 94\ufffd\ufffd239<\/span>94<\/span><\/span><\/span><\/span><\/span><\/span><\/span>P<\/span>u<\/span>239<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> in 1 g is:<\/h3>\n

Number\u00a0of\u00a0moles=MassMolar\u00a0mass=1\u2009g239\u2009g\/mol<\/span>Number\u00a0of\u00a0moles<\/span><\/span>=<\/span><\/span>Molar\u00a0mass<\/span>Mass<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>239g\/mol<\/span>1g<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

Now, we use Avogadro’s number (6.022\u00d71023\u2009atoms\/mol<\/span>6.022<\/span>\u00d7<\/span><\/span>1<\/span>023<\/span><\/span><\/span><\/span><\/span><\/span><\/span>atoms\/mol<\/span><\/span><\/span><\/span><\/span><\/span>) to find the number of atoms:<\/h3>\n

Number\u00a0of\u00a0atoms=Number\u00a0of\u00a0moles\u00d76.022\u00d71023\u2009atoms\/mol<\/span>Number\u00a0of\u00a0atoms<\/span><\/span>=<\/span><\/span>Number\u00a0of\u00a0moles<\/span><\/span>\u00d7<\/span><\/span>6.022<\/span>\u00d7<\/span><\/span>1<\/span>023<\/span><\/span><\/span><\/span><\/span><\/span><\/span>atoms\/mol<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

Once we have the number of atoms, we can find the total energy released:<\/h3>\n

Total\u00a0energy\u00a0released=Number\u00a0of\u00a0atoms\u00d7Energy\u00a0per\u00a0fission<\/span>Total\u00a0energy\u00a0released<\/span><\/span>=<\/span><\/span>Number\u00a0of\u00a0atoms<\/span><\/span>\u00d7<\/span><\/span>Energy\u00a0per\u00a0fission<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

Given that the average energy released per fission is 180 MeV, we can use this value to find the total energy released.<\/h3>\n

Let’s perform the calculations:<\/h3>\n

Number\u00a0of\u00a0moles=1\u2009g239\u2009g\/mol\u22484.18\u00d710\u22123\u2009mol<\/span>Number\u00a0of\u00a0moles<\/span><\/span>=<\/span><\/span>239g\/mol<\/span>1g<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u2248<\/span><\/span>4.18<\/span>\u00d7<\/span><\/span>1<\/span>0\u22123<\/span><\/span><\/span><\/span><\/span><\/span><\/span>mol<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

Number\u00a0of\u00a0atoms=4.18\u00d710\u22123\u00d76.022\u00d71023\u22482.515\u00d71021\u2009atoms<\/span>Number\u00a0of\u00a0atoms<\/span><\/span>=<\/span><\/span>4.18<\/span>\u00d7<\/span><\/span>1<\/span>0\u22123<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>6.022<\/span>\u00d7<\/span><\/span>1<\/span>023<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2248<\/span><\/span>2.515<\/span>\u00d7<\/span><\/span>1<\/span>021<\/span><\/span><\/span><\/span><\/span><\/span><\/span>atoms<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

Total\u00a0energy\u00a0released=2.515\u00d71021\u00d7180\u2009MeV<\/span>Total\u00a0energy\u00a0released<\/span><\/span>=<\/span><\/span>2.515<\/span>\u00d7<\/span><\/span>1<\/span>021<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>180<\/span>MeV<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

Total\u00a0energy\u00a0released\u22484.5287\u00d71023\u2009MeV<\/span>Total\u00a0energy\u00a0released<\/span><\/span>\u2248<\/span><\/span>4.5287<\/span>\u00d7<\/span><\/span>1<\/span>023<\/span><\/span><\/span><\/span><\/span><\/span><\/span>MeV<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

Therefore, if all the atoms in 1 g of pure 94\ufffd\ufffd239<\/span>94<\/span><\/span><\/span><\/span><\/span><\/span><\/span>P<\/span>u<\/span>239<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> undergo fission, approximately 4.5287\u00d71023<\/span>4.5287<\/span>\u00d7<\/span><\/span>1<\/span>023<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> MeV of energy would be released.<\/h3>\n","protected":false},"excerpt":{"rendered":"

The fission properties of Pu239 are very similar to those of 92U235. How much energy (in MeV), is released if all the atoms in 1 g of pure 94Pu239 undergo fission? The average energy released per fission is 180 MeV. Ans. To find the energy released when all the atoms in 1 g of pure […]<\/p>\n","protected":false},"author":21830,"featured_media":114041,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12119],"tags":[],"class_list":{"0":"post-149269","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-quiz"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/149269","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=149269"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/149269\/revisions"}],"predecessor-version":[{"id":149270,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/149269\/revisions\/149270"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/114041"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=149269"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=149269"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=149269"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}