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  • An electron of a hydrogen like atom, having , jumps from <img alt="4^{\text {th }}" src="https://entrancecorner.oncodeco

An electron of a hydrogen like atom, having Z=4, jumps from 4^{\text {th }} energy state to 2^{\text {nd }} energy state. The energy released in this process, will be :
(Given \, \, \mathrm{Rch}=13.6 \mathrm{eV} )
Where \mathrm{R}= Rydberg constant
\mathrm{c}= Speed of light in vacuum
\mathrm{h}= Planck's constant

Option: 1

40.8 \mathrm{eV}


Option: 2

3.4 \mathrm{eV}


Option: 3

10.5 \mathrm{eV}


Option: 4

13.6 \mathrm{eV}


Answers (1)

best_answer

$$ \Delta \mathrm{E}=13.6 \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)
\mathrm{Z}=4 (hydrogen like atom)
$$ \mathrm{n}_1=2, \mathrm{n}_2=4
$$ \begin{aligned} \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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vinayak

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