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The kinetic energy of one electron is `2.8xx10^(-13)` J. What is the de-Broglie wavelengthA. `9.25xx10^(-13)` mB. `9.25xx10^(-16)` mC. `9.25xx10^(-8)` mD. `18.5xx10^(-13)` m |
Answer» Correct Answer - A `because KE=1/2mv^2` Here, `m=9.1xx10^(-31)`kg, KE=`2.8xx10^(-13)`J `therefore 2.8xx10^(-13)=1/2xx9.1xx10^(-31)xxv^2` `v=sqrt((2xx2.8xx10^(-13))/(9.1xx10^(-31)))` `v=0.784xx10^9` m/s `because lambda=h/"mv"=(6.6xx10^(-34))/(9.1xx10^(-31)xx0.784xx10^9)` `=0.925xx10^(-12)` m `=9.25xx10^(-13) m ` |
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