The de Broglie wavelength of electr...
Question

# The de Broglie wavelength of electron of ( mathrm{He}^{+} ) ion is 3.329 A. If the photon emitted upon de-excitation of this ( mathrm{He}^{+} ) ion is made to hit H-atom in its ground state so as to librate electron from it, what will be the de-Broglie's wave length of photoelectron

JEE/Engineering Exams
Chemistry
Solution
1376
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( E=-frac{R_{mathrm{H}}}{n^{2}} )
where
( R_{mathrm{H}}= ) the Rydberg energy constant ( left.2.180 times 10^{-18} mathrm{J}right) )
since ( n=1 )
( E_{1}=-2.180 times 10^{-18} mathrm{J} )
( K E=-E_{1}=2.180 times 10^{-18} mathrm{J} )
( K E=frac{1}{2} m v^{2} )
( v=sqrt{frac{2 K E}{m}} )
( =sqrt{frac{2 times 2.180 times 10^{-18} ot J}{9.11 times 10^{-31} mathrm{kg}} times frac{1 mathrm{kg} cdot mathrm{m}^{2} mathrm{s}^{-2}}{1)^{prime}}} )
( =sqrt{4.786 times 10^{12} mathrm{m}^{2} mathrm{s}^{-2}}=2.188 times 10^{6} mathrm{m} cdot mathrm{s}^{-1} )
Calculate the de Broglie wavelength
( lambda=frac{h}{m v} )
( =frac{6.626 times 10^{-34} text {J.s }}{9.109 times 10^{-31} mathrm{kg} times 2.188 times 10^{6} mathrm{m} cdot mathrm{s}^{-1}} )
( times frac{1 mathrm{kg} cdot mathrm{m}^{2} cdot g^{2}}{1 ot J}=3.325 times 10^{-10} mathrm{m} )
( =332.5 mathrm{pm} )

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