Question 192: Atoms & Nuclei

Question

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

Accepted Answer

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:
$\begin{aligned} \Delta \mathrm{E} & =13.6 \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}ight) \\ \mathrm{Z} & =4 	ext { (hydrogen like atom) } \ \mathrm{n}_1 & =2, \mathrm{n}_2=4 \\ \Delta \mathrm{E} & =13.6(4)^2\left(\frac{1}{4}-\frac{1}{16}ight) \\ & =13.6 	imes\left(\frac{16-4}{64}ight) 	imes 16 \\ \Delta \mathrm{E} & =13.6 	imes \frac{12}{64} 	imes 16 \\ \Delta \mathrm{E} & =40.8 \mathrm { eV}\end{aligned}$

Answer

$$10.5 ~\mathrm{eV}$$

Answer

$$40.8 ~\mathrm{eV}$$

Answer

$$13.6 ~\mathrm{eV}$$

Answer

$$3.4 ~\mathrm{eV}$$

Atoms & Nuclei

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

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