An electron and a proton are separated by a large distance and the electron approaches the proton with K.E. of 2 eV. If the electron is captured by the proton to form a hydrogen atom in the ground state, wavelength of photon emitted will be

$793 \circ A$

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$1096 \circ A$

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$704.5 \circ A$

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$1291.6 \circ A$

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Solution

The correct option is A
$793 \circ A$

Initial energy of electron = 2eV
Energy after formation of hydrogen atom in the ground state
$= - 13.6 e V$ .
Energy released $= 2 - (- 13.6) = 15.6 e V$
$λ = \frac{h c}{15.6} = 793 A$

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An electron and a proton are separated by a large distance and the electron approaches the proton with K.E. of 2 eV. If the electron is captured by the proton to form a hydrogen atom in the ground state, wavelength of photon emitted will be

The correct option is A 793∘A Initial energy of electron = 2eV Energy after formation of hydrogen atom in the ground state =−13.6eV. Energy released =2–(−13.6)=15.6 eV λ=hc15.6=793 A

The correct option is A
$793 \circ A$

Initial energy of electron = 2eV
Energy after formation of hydrogen atom in the ground state
$= - 13.6 e V$ .
Energy released $= 2 - (- 13.6) = 15.6 e V$
$λ = \frac{h c}{15.6} = 793 A$