Atoms & Nuclei

JEE Main level, single correct, medium

52 of 239

Question 52 of 239

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

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Answer: (D) 40.8eV40.8eV

According to Bohr's atomic theory, whenever an electron jumps from one orbit to another, there is always an emission or absorption energy in terms of emission or absorption of photons.
The formula to calculate the energy (∆E)\left(∆E\right) of the emitted or absorbed photon when an electron jumps from a state with quantum number n1{n}_{1} to another with quantum number n2{n}_{2} is given by
∆E=−RchZ2[1ni2−1nf2]....................(1)∆E=-Rch{Z}^{2}\left[\dfrac{1}{{{n}_{i}}^{2}}-\dfrac{1}{{{n}_{f}}^{2}}\right]....................(1)
where, RR is Rydberg's constant, hh is Planck's constant, cc is the speed of light and ZZ is the atomic number.
Substitute the values of the given parameters into equation (1) to calculate the required energy.
∆E=−13.6×(4)2[142−122]eV=−13.6×16×[116−14]eV=40.8eV\begin{matrix}∆E & = & -13.6\times {\left(4\right)}^{2}\left[\dfrac{1}{{4}^{2}}-\dfrac{1}{{2}^{2}}\right]eV \\ & = & -13.6\times 16\times \left[\dfrac{1}{16}-\dfrac{1}{4}\right]eV \\ & = & 40.8eV\end{matrix}

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