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If the kinetic energy of electron increases by 44% find the percentage change in De-Brogile's wavelength
Asked by punitkumar8728 | 06 Apr, 2021, 21:17: PM
answered-by-expert Expert Answer
Relation between De Broglie Wavelength and Kinetic Energy-
lambda equals fraction numerator h over denominator square root of 2 m K end root end fraction m space a n d space h space a r e space c o n s tan t space f o r space a space e l e c t r o n. S o comma space lambda alpha fraction numerator 1 over denominator square root of K end fraction lambda subscript 1 over lambda subscript 2 equals fraction numerator square root of 1.44 K end root over denominator square root of K end fraction lambda subscript 2 equals fraction numerator lambda over denominator 1.2 end fraction increment lambda equals lambda minus fraction numerator lambda over denominator 1.2 end fraction increment lambda equals fraction numerator 0.2 lambda over denominator 1.2 end fraction fraction numerator increment lambda over denominator lambda end fraction cross times 100 equals 2 over 12 cross times 100 equals 100 over 6 equals 16.7 percent sign
Wavelength changes by 16.7%
Answered by Ravi | 15 Apr, 2021, 20:14: PM

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