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If kinetic energy of emitted electron is made double then its de-Broglie wavelength will become……times initial de-Broglie wavelength. |
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Answer» `sqrt(2)` but `p^(2)=sqrt(2mK)` `therefore lambda =(h)/(sqrt(2mK))` `therefore lambda prop (1)/(sqrt(K))`[`because(h)/(sqrt(2m))` same ] `therefore (lambda_(2))/(lambda_(1))=sqrt((K_(1))/(K_(2)))` `therefore (lambda_(2))/(lambda_(1))=sqrt((K)/(2K)) therefore (lambda_(2))/(lambda_(1))=(1)/(sqrt(2))` |
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