JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 18)
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
$$10.5 ~\mathrm{eV}$$
$$40.8 ~\mathrm{eV}$$
$$13.6 ~\mathrm{eV}$$
$$3.4 ~\mathrm{eV}$$
Explanation
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:
$\begin{aligned} \Delta \mathrm{E} & =13.6 \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\\\ \mathrm{Z} & =4 \text { (hydrogen like atom) } \\ \mathrm{n}_1 & =2, \mathrm{n}_2=4 \\\\ \Delta \mathrm{E} & =13.6(4)^2\left(\frac{1}{4}-\frac{1}{16}\right) \\\\ & =13.6 \times\left(\frac{16-4}{64}\right) \times 16 \\\\ \Delta \mathrm{E} & =13.6 \times \frac{12}{64} \times 16 \\\\ \Delta \mathrm{E} & =40.8 \mathrm { eV}\end{aligned}$
$\begin{aligned} \Delta \mathrm{E} & =13.6 \mathrm{Z}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\\\ \mathrm{Z} & =4 \text { (hydrogen like atom) } \\ \mathrm{n}_1 & =2, \mathrm{n}_2=4 \\\\ \Delta \mathrm{E} & =13.6(4)^2\left(\frac{1}{4}-\frac{1}{16}\right) \\\\ & =13.6 \times\left(\frac{16-4}{64}\right) \times 16 \\\\ \Delta \mathrm{E} & =13.6 \times \frac{12}{64} \times 16 \\\\ \Delta \mathrm{E} & =40.8 \mathrm { eV}\end{aligned}$
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