{"id":87509,"date":"2025-06-01T06:45:42","date_gmt":"2025-06-01T06:45:42","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87509"},"modified":"2025-06-01T06:45:42","modified_gmt":"2025-06-01T06:45:42","slug":"if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/","title":{"rendered":"If the potential difference applied to an X-ray tube is doubled while"},"content":{"rendered":"

If the potential difference applied to an X-ray tube is doubled while keeping the separation between the filament and the target as same, what will happen to the cutoff wavelength?<\/p>\n

[amp_mcq option1=”Will remain same” option2=”Will be doubled” option3=”Will be halved” option4=”Will be four times of the original wavelength” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2017<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe question asks what happens to the cutoff wavelength of X-rays produced in an X-ray tube if the potential difference is doubled while keeping the filament-target separation the same.
\n<\/section>\n
\nThe production of X-rays in an X-ray tube involves accelerating electrons through a potential difference ($V$) and making them strike a target. The energy gained by an electron accelerated through potential $V$ is $eV$, where $e$ is the elementary charge. This energy is converted into electromagnetic radiation (X-rays) and heat. The minimum wavelength ($\\lambda_{min}$) of the emitted X-rays, also known as the cutoff wavelength or Duane-Hunt limit, occurs when the entire energy of the electron is converted into a single X-ray photon. The energy of a photon is $hf = hc\/\\lambda$, where $h$ is Planck’s constant, $f$ is frequency, $c$ is the speed of light, and $\\lambda$ is the wavelength.
\nAccording to the Duane-Hunt law: $eV = hc\/\\lambda_{min}$
\nThis equation shows that the cutoff wavelength $\\lambda_{min}$ is inversely proportional to the applied potential difference $V$: $\\lambda_{min} \\propto 1\/V$.
\nIf the potential difference is doubled, i.e., $V’ = 2V$, the new cutoff wavelength $\\lambda’_{min}$ will be:
\n$\\lambda’_{min} = \\frac{hc}{e(2V)} = \\frac{1}{2} \\left(\\frac{hc}{eV}\\right) = \\frac{1}{2} \\lambda_{min}$
\nSo, the cutoff wavelength will be halved. The separation between the filament and the target does not directly affect the cutoff wavelength, which is determined by the maximum energy of the electrons reaching the target, dictated by the potential difference.
\n<\/section>\n
\nThe X-ray spectrum produced consists of a continuous spectrum (bremsstrahlung) with a minimum wavelength $\\lambda_{min}$ and characteristic peaks at specific wavelengths. The cutoff wavelength depends only on the accelerating voltage, while the intensity and characteristic peaks also depend on the target material and the electron beam current.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

If the potential difference applied to an X-ray tube is doubled while keeping the separation between the filament and the target as same, what will happen to the cutoff wavelength? [amp_mcq option1=”Will remain same” option2=”Will be doubled” option3=”Will be halved” option4=”Will be four times of the original wavelength” correct=”option3″] This question was previously asked in … <\/p>\n

Detailed SolutionIf the potential difference applied to an X-ray tube is doubled while<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1093],"tags":[1101,1128,1274],"class_list":["post-87509","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1101","tag-physics","tag-wave-motion","no-featured-image-padding"],"yoast_head":"\nIf the potential difference applied to an X-ray tube is doubled while<\/title>\n<meta name=\"description\" content=\"The question asks what happens to the cutoff wavelength of X-rays produced in an X-ray tube if the potential difference is doubled while keeping the filament-target separation the same. The production of X-rays in an X-ray tube involves accelerating electrons through a potential difference ($V$) and making them strike a target. The energy gained by an electron accelerated through potential $V$ is $eV$, where $e$ is the elementary charge. This energy is converted into electromagnetic radiation (X-rays) and heat. The minimum wavelength ($lambda_{min}$) of the emitted X-rays, also known as the cutoff wavelength or Duane-Hunt limit, occurs when the entire energy of the electron is converted into a single X-ray photon. The energy of a photon is $hf = hc\/lambda$, where $h$ is Planck's constant, $f$ is frequency, $c$ is the speed of light, and $lambda$ is the wavelength. According to the Duane-Hunt law: $eV = hc\/lambda_{min}$ This equation shows that the cutoff wavelength $lambda_{min}$ is inversely proportional to the applied potential difference $V$: $lambda_{min} propto 1\/V$. If the potential difference is doubled, i.e., $V' = 2V$, the new cutoff wavelength $lambda'_{min}$ will be: $lambda'_{min} = frac{hc}{e(2V)} = frac{1}{2} left(frac{hc}{eV}right) = frac{1}{2} lambda_{min}$ So, the cutoff wavelength will be halved. The separation between the filament and the target does not directly affect the cutoff wavelength, which is determined by the maximum energy of the electrons reaching the target, dictated by the potential difference.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"If the potential difference applied to an X-ray tube is doubled while\" \/>\n<meta property=\"og:description\" content=\"The question asks what happens to the cutoff wavelength of X-rays produced in an X-ray tube if the potential difference is doubled while keeping the filament-target separation the same. The production of X-rays in an X-ray tube involves accelerating electrons through a potential difference ($V$) and making them strike a target. The energy gained by an electron accelerated through potential $V$ is $eV$, where $e$ is the elementary charge. This energy is converted into electromagnetic radiation (X-rays) and heat. The minimum wavelength ($lambda_{min}$) of the emitted X-rays, also known as the cutoff wavelength or Duane-Hunt limit, occurs when the entire energy of the electron is converted into a single X-ray photon. The energy of a photon is $hf = hc\/lambda$, where $h$ is Planck's constant, $f$ is frequency, $c$ is the speed of light, and $lambda$ is the wavelength. According to the Duane-Hunt law: $eV = hc\/lambda_{min}$ This equation shows that the cutoff wavelength $lambda_{min}$ is inversely proportional to the applied potential difference $V$: $lambda_{min} propto 1\/V$. If the potential difference is doubled, i.e., $V' = 2V$, the new cutoff wavelength $lambda'_{min}$ will be: $lambda'_{min} = frac{hc}{e(2V)} = frac{1}{2} left(frac{hc}{eV}right) = frac{1}{2} lambda_{min}$ So, the cutoff wavelength will be halved. The separation between the filament and the target does not directly affect the cutoff wavelength, which is determined by the maximum energy of the electrons reaching the target, dictated by the potential difference.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:45:42+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"If the potential difference applied to an X-ray tube is doubled while","description":"The question asks what happens to the cutoff wavelength of X-rays produced in an X-ray tube if the potential difference is doubled while keeping the filament-target separation the same. The production of X-rays in an X-ray tube involves accelerating electrons through a potential difference ($V$) and making them strike a target. The energy gained by an electron accelerated through potential $V$ is $eV$, where $e$ is the elementary charge. This energy is converted into electromagnetic radiation (X-rays) and heat. The minimum wavelength ($lambda_{min}$) of the emitted X-rays, also known as the cutoff wavelength or Duane-Hunt limit, occurs when the entire energy of the electron is converted into a single X-ray photon. The energy of a photon is $hf = hc\/lambda$, where $h$ is Planck's constant, $f$ is frequency, $c$ is the speed of light, and $lambda$ is the wavelength. According to the Duane-Hunt law: $eV = hc\/lambda_{min}$ This equation shows that the cutoff wavelength $lambda_{min}$ is inversely proportional to the applied potential difference $V$: $lambda_{min} propto 1\/V$. If the potential difference is doubled, i.e., $V' = 2V$, the new cutoff wavelength $lambda'_{min}$ will be: $lambda'_{min} = frac{hc}{e(2V)} = frac{1}{2} left(frac{hc}{eV}right) = frac{1}{2} lambda_{min}$ So, the cutoff wavelength will be halved. The separation between the filament and the target does not directly affect the cutoff wavelength, which is determined by the maximum energy of the electrons reaching the target, dictated by the potential difference.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/","og_locale":"en_US","og_type":"article","og_title":"If the potential difference applied to an X-ray tube is doubled while","og_description":"The question asks what happens to the cutoff wavelength of X-rays produced in an X-ray tube if the potential difference is doubled while keeping the filament-target separation the same. The production of X-rays in an X-ray tube involves accelerating electrons through a potential difference ($V$) and making them strike a target. The energy gained by an electron accelerated through potential $V$ is $eV$, where $e$ is the elementary charge. This energy is converted into electromagnetic radiation (X-rays) and heat. The minimum wavelength ($lambda_{min}$) of the emitted X-rays, also known as the cutoff wavelength or Duane-Hunt limit, occurs when the entire energy of the electron is converted into a single X-ray photon. The energy of a photon is $hf = hc\/lambda$, where $h$ is Planck's constant, $f$ is frequency, $c$ is the speed of light, and $lambda$ is the wavelength. According to the Duane-Hunt law: $eV = hc\/lambda_{min}$ This equation shows that the cutoff wavelength $lambda_{min}$ is inversely proportional to the applied potential difference $V$: $lambda_{min} propto 1\/V$. If the potential difference is doubled, i.e., $V' = 2V$, the new cutoff wavelength $lambda'_{min}$ will be: $lambda'_{min} = frac{hc}{e(2V)} = frac{1}{2} left(frac{hc}{eV}right) = frac{1}{2} lambda_{min}$ So, the cutoff wavelength will be halved. The separation between the filament and the target does not directly affect the cutoff wavelength, which is determined by the maximum energy of the electrons reaching the target, dictated by the potential difference.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:45:42+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/","url":"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/","name":"If the potential difference applied to an X-ray tube is doubled while","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:45:42+00:00","dateModified":"2025-06-01T06:45:42+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The question asks what happens to the cutoff wavelength of X-rays produced in an X-ray tube if the potential difference is doubled while keeping the filament-target separation the same. The production of X-rays in an X-ray tube involves accelerating electrons through a potential difference ($V$) and making them strike a target. The energy gained by an electron accelerated through potential $V$ is $eV$, where $e$ is the elementary charge. This energy is converted into electromagnetic radiation (X-rays) and heat. The minimum wavelength ($\\lambda_{min}$) of the emitted X-rays, also known as the cutoff wavelength or Duane-Hunt limit, occurs when the entire energy of the electron is converted into a single X-ray photon. The energy of a photon is $hf = hc\/\\lambda$, where $h$ is Planck's constant, $f$ is frequency, $c$ is the speed of light, and $\\lambda$ is the wavelength. According to the Duane-Hunt law: $eV = hc\/\\lambda_{min}$ This equation shows that the cutoff wavelength $\\lambda_{min}$ is inversely proportional to the applied potential difference $V$: $\\lambda_{min} \\propto 1\/V$. If the potential difference is doubled, i.e., $V' = 2V$, the new cutoff wavelength $\\lambda'_{min}$ will be: $\\lambda'_{min} = \\frac{hc}{e(2V)} = \\frac{1}{2} \\left(\\frac{hc}{eV}\\right) = \\frac{1}{2} \\lambda_{min}$ So, the cutoff wavelength will be halved. The separation between the filament and the target does not directly affect the cutoff wavelength, which is determined by the maximum energy of the electrons reaching the target, dictated by the potential difference.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/if-the-potential-difference-applied-to-an-x-ray-tube-is-doubled-while\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"If the potential difference applied to an X-ray tube is doubled while"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87509","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=87509"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87509\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87509"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87509"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87509"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}