Verified Solution • 2025–26 Session moderate

QJEEM13Phc49d: JEE Main 2013 Physics - Class 12 JEEM13Phc49d Solution: In a hydrogen like atom electron make transition from an energy …

In a hydrogen like atom electron make transition from an energy level with quantum number nn to another with quantum number (n−1)\left( {n - 1} \right) . If n > > 1, the frequency of radiation emitted is proportional to :

Multiple Choice Options

  • A 1n{1 \over n}
  • B 1n2{1 \over {{n^2}}}
  • C 1n32{1 \over {{n^{{3 \over 2}}}}}
  • D 1n3{1 \over {{n^3}}}

Step-by-Step Verified Solution

Δ\Delta E =13.6Z2[1n12−1n22]= 13.6{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]

As Δ\Delta E = h υ\upsilon

⇒\Rightarrow Δ\Delta E ∝\propto υ\upsilon ∝\propto [1n12−1n22]\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]

Here n1 = n - 1 and n2 = n

∴\therefore υ\upsilon ∝\propto [1(n−1)2−1n2]\left[ {{1 \over {{{\left( {n - 1} \right)}^2}}} - {1 \over {{n^2}}}} \right]

⇒\Rightarrow υ\upsilon ∝\propto [2n−1n2(n−1)2]\left[ {{{2n - 1} \over {{n^2}{{\left( {n - 1} \right)}^2}}}} \right]

[ As n > > 1, so we can assume n - 1 = n and 2n - 1 = 2n ]

⇒\Rightarrow υ\upsilon ∝\propto [2nn2(n)2]\left[ {{{2n} \over {{n^2}{{\left( {n } \right)}^2}}}} \right] ∝\propto [2nn4]\left[ {{{2n} \over {{n^4}}}} \right] ∝\propto [2n3]\left[ {{2 \over {{n^3}}}} \right]

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