Energy E of a hydrogen atom with principal quantum number n is given by E = n 2 − 13.6 eV . The energy of a photon ejected when the electron jumps for n = 3 state n = 2 state of hydrogen is approximatelyNEET 2004

Energy of photon = E 2 − E 2 = 9 13.6 − ( 9 − 13.6 ) = 30 5 × 13.6 = 1.9 eV

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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2004

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Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{- 13.6}{n ^{2}} eV$ . The energy of a photon ejected when the electron jumps for $n = 3$ state $n = 2$ state of hydrogen is approximatelyNEET 2004

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Solution:

Energy of photon $= E_{2} - E_{2}$ $= \frac{13.6}{9} - (\frac{- 13.6}{9}) = \frac{5}{30} \times 13.6 = 1.9 eV$

Stream:NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2004

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