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The de Broglie wavelength of electron of He+ ion is 3.32 Å. If the photon emitted upon de-excitation of this He+ ion is made to hit H-atom in its ground state so as to liberate electron from it, then the de Broglie wavelength of photoelectron is:

A
2.348 Å
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B
1.917 Å
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C
3.329 Å
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D
3.329 pm
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Solution

The correct option is A 2.348 Å
Energy of the photon liberated:
Using formula E=hcλ
Where h = planck’s constant 6.626×1034 J s
c = velocity of light = 3×108 m s1
and λ is the wavelength (given) = 3.32 Å=3.32×1010 m
Putting all these values in the formula:
E=6.626×1034×3×1083.32×1010

= 5.947×1016J
Now using formula for de-Broglie wavelength i.e., λ=hm×E
m = mass of electron = 9.109×1031 kg
λ=6.626×1034(9.109×1031×5.947×1016)

= 6.626×10345.4171×1046

= 2.348 Å


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