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Question

13.6 eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum energy required for it to escape from the atom. If the wavelength of emitted electron is x oA, find the nearest integer value of x
(me=9.1×1031 kg, h=6.626×1034 Js)

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Solution

Given, ionisation energy = 13.6 eV
Energy absorbed by the electron = 13.6×1.5=20.4 eV
So, the maximum kinetic energy of the electron
=20.413.6 eV=6.8 eV

Again, from de Broglie wavelength,
λ=h2meV
λ=6.626×10342×6.8×1.6×1019×9.1×1031
λ=6.626×103414×1025
λ=4.73 Ao5 oA

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