{"id":17058,"date":"2025-06-12T07:33:11","date_gmt":"2025-06-12T07:33:11","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=17058"},"modified":"2025-06-12T07:33:16","modified_gmt":"2025-06-12T07:33:16","slug":"star-a-has-a-surface-temperature-of-8000k-and-its-thermal-emission-peaks-around-360nm","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/star-a-has-a-surface-temperature-of-8000k-and-its-thermal-emission-peaks-around-360nm\/","title":{"rendered":"Star A has a surface temperature of 8000K and its thermal emission peaks around 360nm"},"content":{"rendered":"\n<p>Star A has a surface temperature of 8000K and its thermal emission peaks around 360nm. Star B is twice as cold, with a temperature of 4000K. What wavelength does its curve peak?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>To find the peak wavelength of thermal emission for a star, we use <strong>Wien&#8217;s Displacement Law<\/strong>, which states: \u03bbmax=bT\\lambda_{\\text{max}} = \\frac{b}{T}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03bbmax\\lambda_{\\text{max}} is the peak wavelength in <strong>meters<\/strong>,<\/li>\n\n\n\n<li>TT is the surface temperature in <strong>Kelvin (K)<\/strong>,<\/li>\n\n\n\n<li>bb is Wien\u2019s displacement constant, approximately <strong>2.898 \u00d7 10\u207b\u00b3 m\u00b7K<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Apply Wien\u2019s Law to Star B<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Temperature of Star B = 4000 K<\/li>\n\n\n\n<li>b=2.898\u00d710\u22123\u2009m\\cdotpKb = 2.898 \\times 10^{-3} \\, \\text{m\u00b7K}<\/li>\n<\/ul>\n\n\n\n<p>\u03bbmax,&nbsp;B=2.898\u00d710\u221234000=7.245\u00d710\u22127\u2009m\\lambda_{\\text{max, B}} = \\frac{2.898 \\times 10^{-3}}{4000} = 7.245 \\times 10^{-7} \\, \\text{m}<\/p>\n\n\n\n<p>Convert to <strong>nanometers (nm)<\/strong>: 7.245\u00d710\u22127\u2009m=724.5\u2009nm7.245 \\times 10^{-7} \\, \\text{m} = 724.5 \\, \\text{nm}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer: The peak wavelength of Star B is approximately 724.5 nanometers (nm).<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (\u2248300 words):<\/h3>\n\n\n\n<p>Wien&#8217;s Displacement Law explains how the temperature of a blackbody (like a star) determines the <strong>wavelength<\/strong> at which it emits most of its radiation. This law shows an <strong>inverse relationship<\/strong> between temperature and peak wavelength: as the temperature increases, the peak wavelength shifts to <strong>shorter<\/strong> (bluer) wavelengths; as the temperature decreases, the peak shifts to <strong>longer<\/strong> (redder) wavelengths.<\/p>\n\n\n\n<p>In this problem, Star A has a surface temperature of 8000 K and peaks at 360 nm, which is in the <strong>ultraviolet<\/strong> part of the spectrum. Star B is <strong>twice as cool<\/strong>, at 4000 K. Since the temperature is halved, the peak wavelength should <strong>double<\/strong>, which gives a good estimate even before calculating.<\/p>\n\n\n\n<p>Using Wien\u2019s Law: \u03bbmax=2.898\u00d710\u221234000=724.5\u2009nm\\lambda_{\\text{max}} = \\frac{2.898 \\times 10^{-3}}{4000} = 724.5 \\, \\text{nm}<\/p>\n\n\n\n<p>This places Star B\u2019s peak in the <strong>red\/infrared<\/strong> part of the spectrum. That means Star B appears <strong>redder<\/strong> than Star A.<\/p>\n\n\n\n<p>This concept is important in astronomy because it helps classify stars based on color and temperature. Hotter stars (blue or white) emit more energy at shorter wavelengths, while cooler stars (red or orange) emit more at longer wavelengths.<\/p>\n\n\n\n<p>Understanding these temperature-wavelength relationships helps astronomers determine a star\u2019s properties by just analyzing its spectrum.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Star A has a surface temperature of 8000K and its thermal emission peaks around 360nm. Star B is twice as cold, with a temperature of 4000K. What wavelength does its curve peak? The correct answer and explanation is: To find the peak wavelength of thermal emission for a star, we use Wien&#8217;s Displacement Law, which [&hellip;]<\/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-17058","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/17058","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=17058"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/17058\/revisions"}],"predecessor-version":[{"id":17059,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/17058\/revisions\/17059"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=17058"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=17058"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=17058"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}