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The energy that should be added to an electron, to reduce its de-Broglie wavelength from `2xx10^(-9) m` to `0.5xx10^(-9)m` will be:A. 1.1 MeVB. 0.56 MeVC. 0.56KeVD. 5.6 eV |
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Answer» `lambda=h/p=h/(sqrt(2mE))` `:. E=(h^(2))/(2m lambda^(2))` `Delta E=(h^(2))/(2m)(1/(lambda_(1)^(2))-1/(lambda_(2)^(2)))` Put `lambda_(1)=0/5xx10^(-9) m` & `lambda_(2)=2xx10^(-9) m` and solve. |
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