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The kinetic energy of a beam of electrons, accelerated through a potential V, equals the energy of a photon of wavelength 5460 nm. Find the de-Broglie wavelength associated with this beam of electrons. |
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Answer» Solution :As per question, kinetic enegy of electron K=energy of a photon of wavelength `lamda=5460nm=5460xx10^(-9)m ` THUS, `K=(HC)/(lamda)=(6.63xx10^(-34)xx3xx10^(8))/(5460xx10^(-9))=3.64xx10^(-20)J` `therefore`de-Broglie wavelength `lamda_(de)=(H)/(sqrt(2mK))=(6.63xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx3.64xx10^(-20)))=2.62xx10^(-9)m=2.62m`. |
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