Ionization potential of hydrogen atom is 13.6 | vedclass

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(c) Final energy of electron $ =  - 13.6 + 12.1 =  - 1.51eV$ which is corresponds to third level $i.e.\,\,n=3$.

Hence number of spectral

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(c) Final energy of electron $ =  - 13.6 + 12.1 =  - 1.51eV$ which is corresponds to third level $i.e.\,\,n=3$.

Hence number of spectral lines emitted   $=\frac{n(n-1)}{2}=\frac{3(3-1)}{2}=3$

Ionization potential of hydrogen atom is $13.6 V$. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy $12.1 eV.$ The spectral lines emitted by hydrogen atoms according to Bohr's theory will be

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