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Questions  >>  JEEMAIN and NEET  >>  NEET PAST PAPERS  >>  2004
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Q)

Energy E of a hydrogen atom with principal quantum number n is given by $E=\frac{-13.6} {n^{6}}eV$. n The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately :-


( A ) 3.4 eV
( B ) 1.9 eV
( C ) 1.5 eV
( D ) 0. 85 eV

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