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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. What is the wavelength and velocity of the emitted electron?

A
v=4.0×106 m/s;λ=4.7×1010 m
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B
v=1.55×106 m/s;λ=4.7×1010 m
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C
v=1.55×106 m/s;λ=16.0×1012 m
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D
v=4.0×106 m/s;λ=16.0×1012 m
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Solution

The correct option is B v=1.55×106 m/s;λ=4.7×1010 m
1.5 times of 13.6 eV, i.e., 20.4 eV, is absorbed by the hydrogen atom out of which 6.8 eV (20.4 - 13.6) is convereted to kinetic energy. KE=6.8 eV=6.8×(1.602×1019coulomb)×(1volt)=1.09×1018 J
Now, KE=12mv2
or v=2KEm= 2(1.09×1018 J(9.1×1031 kg)=1.55×106 m/s.
From De Broglie equation we have :
λ=hmv=(6.63×1034J.s)(9.1×1031 kg)(1.55×106 m/s)
=4.70×1010 m

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