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Question

# An electron is in an excited state in a hydrogen like atom. It has a total energy of −3.4 eV. The kinetic energy of the electron is E and its de Broglie wavelength is λ.

A
E=6.8 eV, λ=6.6×1010 m
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B
E=3.4 eV, λ=6.6×1010 m
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C
E=3.4 eV, λ=6.6×1011 m
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D
E=6.8 eV, λ=6.6×1011 m
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Solution

## The correct option is B E=3.4 eV, λ=−6.6×10−10 mThe potential energy of an electron is twice the kinetic energy(with negative sign) of an electron.PE=−2EThe total energy is:TE=PE+KE−3.4=−2E+EE=3.4eVLet, p be the momentum of electron and m be the mass of electron.E=p22mp=√2mENow, the De-Broglie wavelength associated with an electron isλ=hpλ=h√2mEλ=6.6×10−34√2×9.1×10−31×(−3.4)×1.6×10−19λ=−6.6×10−349.95×10−25λ=−0.66×10−9λ=−6.6×10−10m

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