Verified Solution • 2025–26 Session moderate

QJEEM23Ph0462: JEE Main 2023 Physics - Class 12 JEEM23Ph0462 Solution: An electron of a hydrogen like atom, having , jumps from energy …

An electron of a hydrogen like atom, having Z=4Z=4 , jumps from 4th 4^{\text {th }} energy state to 2nd 2^{\text {nd }} energy state. The energy released in this process, will be :

(Given Rch = 13.6 eV13.6~\mathrm{eV} )

Where R = Rydberg constant

c = Speed of light in vacuum

h = Planck's constant

Multiple Choice Options

  • A 10.5 eV10.5 ~\mathrm{eV}
  • B 40.8 eV40.8 ~\mathrm{eV}
  • C 13.6 eV13.6 ~\mathrm{eV}
  • D 3.4 eV3.4 ~\mathrm{eV}

Step-by-Step Verified Solution

The energy difference between the 4th and 2nd energy states of a hydrogen-like atom can be calculated using the formula for the energy levels of a hydrogen-like atom:

ΔEamp;=13.6Z2(1n12−1n22)Zamp;=4 (hydrogen like atom) n1amp;=2,n2=4ΔEamp;=13.6(4)2(14−116)amp;=13.6×(16−464)×16ΔEamp;=13.6×1264×16ΔEamp;=40.8eV\begin{aligned} \Delta \mathrm{E} & =13.6 \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\\\ \mathrm{Z} & =4 \text { (hydrogen like atom) } \\ \mathrm{n}_1 & =2, \mathrm{n}_2=4 \\\\ \Delta \mathrm{E} & =13.6(4)^2\left(\frac{1}{4}-\frac{1}{16}\right) \\\\ & =13.6 \times\left(\frac{16-4}{64}\right) \times 16 \\\\ \Delta \mathrm{E} & =13.6 \times \frac{12}{64} \times 16 \\\\ \Delta \mathrm{E} & =40.8 \mathrm { eV}\end{aligned}

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