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$L{i^{2 + }}$ ion is in the excited state with wave function of its electron being denoted by the symbol ${\Psi _{3,1,0}}$. How many radiations of different wavelengths will this $L{i^{2 + }}$ ion will emit upon de-excitation?
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A. zero
B. $6$
C. $3$
D. $1$

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Last updated date: 20th Jun 2024
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Answer
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Hint: $L{i^{2 + }}$ ion has $2$ electrons removed from its orbits, so it behaves analogous to a hydrogen atom. The subscripts in the wave functions indicate the different quantum numbers of the electron. Out of this, only the principal quantum number is needed to find the number of radiations emitted.

Formulas used: number of spectral lines emitted when an electron jumps from excited state to ground state $ = \dfrac{{n(n - 1)}}{2}$
Where $n$ denotes the principal quantum number

Complete step by step answer:
As we know the lithium atom has a total of $3$ electrons. Therefore, the $L{i^{2 + }}$ ion has $2$ electrons removed from the three and has only one electron, which is present in the first orbital ($1s$). Thus, it is similar to the hydrogen atom in its ground state.
The subscripts in the wave functions indicate the different quantum numbers of the electron, namely the principal quantum number, azimuthal quantum number and the magnetic quantum number respectively. To find the number of radiations emitted (which is equal to the number of spectral lines produced), we only need to know the principal quantum number. The principal quantum number is the first number mentioned in the wave function and gives us the shell to which the electron belongs to. Hence, in this case, as the wave function is ${\Psi _{3,1,0}}$, the electron belongs to the third orbit.
From the Bohr model of the atom,
The formula for finding number of spectral lines emitted when an electron jumps from excited state to ground state $ = \dfrac{{n(n - 1)}}{2}$
Where $n$ denotes the principal quantum number.
Therefore, substituting our value as $n = 3$, we get:
Number of radiations $ = \dfrac{{3(3 - 1)}}{2} = \dfrac{{3 \times 2}}{2} = 3$

So, the correct answer is Option C.

Note: The radiations emitted in this case from an emission spectrum, and thus, can be viewed as thin bright lies on an otherwise dark background. The wavelength and frequency of the emitted radiations is dependent on the energy difference between the two orbits. The azimuthal quantum number ($l$) denotes the orbital to which the electron belongs to. The values of $l$ for different orbitals are: $s = 0,p = 1,d = 2,f = 3$. The magnetic quantum number describes the orientation of the region in space occupied by an electron with respect to an applied magnetic field, is denoted by ${m_l}$ and can have values from $ - l$ to $ + l$. Also note that we used the $\dfrac{{n(n - 1)}}{2}$ formula only because the $L{i^{2 + }}$ ion resembles the hydrogen atom in its ground state.