Since comments caused certain level of confusion, I guess I'll try to provide a further illustration. You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$. Electron doesn't get stuck forever on any of the levels with $n > 1$. Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms. In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$. The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this. I put together a rough drawing in Inkscape to illustrate all possible transitions*: I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots). To determine $N$, you need to sum the states, as Soumik Das rightfully commented: $$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$ For $n = 6$: $$N = \frac{6(6-1)}{2} = 15$$ Obviously the same result is obtained by taking the sum directly. * Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$. Answer Verified Hint: When an electron absorbs energy it becomes excited and jumps from lower energy level to higher energy level and when it comes back to its ground state, the same amount of energy is released as was absorbed when it went to higher energy level. Complete answer: When an electron in a higher energy state jumps to a lower energy state in Bohr’s model of hydrogen atom, it emits a photon which has energy equal to the difference in the energy between the final and initial energy states.\[h\nu = {E_f} - {E_i}\]Where, \[\nu \]= frequency of photon emittedEach photon emitted corresponds to an emission line in the spectrum.So, an electron can make the transition from a higher energy state to any lower energy state provided it emits a photon given by the above equation. In the question given, if an electron is in a higher energy state; the maximum number of spectral lines is when the electron jumps to the next lowest energy state and so on as then photons will be emitted during each transition.When an electron in the first energy level of a Hydrogen atom absorbs energy, it becomes excited and jumps to \[{6^{th}}\] energy level. When it returns back, it can return straight back to the first energy level or in several steps. The following transitions may be possible, that is,It can jump from\[6 \to 5,6 \to 4,6 \to 3,6 \to 2,6 \to 1[5steps]\]or from \[6 \to 4,6 \to 3,6 \to 2,6 \to 1\] (4 steps)and so on till it reaches the single step transitionThat is, from \[6 \to 1\]Hence, total lines obtained on the emission spectrum will be\[(5 + 4 + 3 + 2 + 1) = 15lines\]The number of spectral lines can also be found by the formula given by Neil Bohr as follows:According to Bohr’s Model of Hydrogen spectrum,The number of spectral lines formed when electron jump or drop from the \[{n^{th}}\]level to ground level = \[\dfrac{{n(n - 1)}}{2}\] here \[n = 6\]So, \[\dfrac{{6(6 - 1)}}{2} = 15\]Hence, the answer is option B which is 15 is correct. Note: Remember that an excited electron can return back directly and also in several steps. However, once it returns to its ground state, the energy released will be the same. Try the new Google Books Check out the new look and enjoy easier access to your favorite features Maximum number of spectral lines emitted when electrons jump from n=5 to n=1 in hydrogen atom sample is 1) 3s 2) 2p 3) 2s 4) 1s Open in App 1Since comments caused certain level of confusion, I guess I'll try to provide a further illustration. You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$. Electron doesn't get stuck forever on any of the levels with $n > 1$. Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms. In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$. The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this. I put together a rough drawing in Inkscape to illustrate all possible transitions*: I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots). To determine $N$, you need to sum the states, as Soumik Das rightfully commented: $$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$ For $n = 6$: $$N = \frac{6(6-1)}{2} = 15$$ Obviously the same result is obtained by taking the sum directly. * Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$. Answer Verified Hint: There is a formula to calculate the number of spectral lines or emission lines formed by the electrons when it drops from orbit ‘n’ to ground state and it is as follows.\[\text{Number of spectral lines = }\dfrac{n(n-1)}{2}\] Where n = orbit from where the electron starts to drop to ground state. Complete step by step answer: - In the question it is given that an electron of hydrogen atom drops from n = 6 to ground state.- We have to calculate the number of maximum spectral lines or emission lines that are going to be obtained when the electron drops from n = 6 to ground state.- Means here n = 6.- Substitute n value in the below formula to get the number of spectral lines formed.\[\begin{align} & \text{Number of spectral lines = }\dfrac{n(n-1)}{2} \\ & =\dfrac{6(6-1)}{2} \\ & =15 \\ \end{align}\] - Therefore the maximum number of emission lines formed when the excited electron of H atom in n = 6 drops to the ground state is 15.- Means an electron forms 15 emission lines when it drops from n = 6 to ground level. Note: When the electron drops from higher orbit to lower orbit means the electron losing or emitting energy then the electron forms some lines related to emitted energy then they are called as emission lines or spectral lines. When an electron jumps to a higher energy level then it absorbs some energy and forms an absorption spectrum or absorption spectral lines. Answer Verified Hint: When an electron absorbs energy it becomes excited and jumps from lower energy level to higher energy level and when it comes back to its ground state, the same amount of energy is released as was absorbed when it went to higher energy level. Complete answer: When an electron in a higher energy state jumps to a lower energy state in Bohr’s model of hydrogen atom, it emits a photon which has energy equal to the difference in the energy between the final and initial energy states.\[h\nu = {E_f} - {E_i}\]Where, \[\nu \]= frequency of photon emittedEach photon emitted corresponds to an emission line in the spectrum.So, an electron can make the transition from a higher energy state to any lower energy state provided it emits a photon given by the above equation. In the question given, if an electron is in a higher energy state; the maximum number of spectral lines is when the electron jumps to the next lowest energy state and so on as then photons will be emitted during each transition.When an electron in the first energy level of a Hydrogen atom absorbs energy, it becomes excited and jumps to \[{6^{th}}\] energy level. When it returns back, it can return straight back to the first energy level or in several steps. The following transitions may be possible, that is,It can jump from\[6 \to 5,6 \to 4,6 \to 3,6 \to 2,6 \to 1[5steps]\]or from \[6 \to 4,6 \to 3,6 \to 2,6 \to 1\] (4 steps)and so on till it reaches the single step transitionThat is, from \[6 \to 1\]Hence, total lines obtained on the emission spectrum will be\[(5 + 4 + 3 + 2 + 1) = 15lines\]The number of spectral lines can also be found by the formula given by Neil Bohr as follows:According to Bohr’s Model of Hydrogen spectrum,The number of spectral lines formed when electron jump or drop from the \[{n^{th}}\]level to ground level = \[\dfrac{{n(n - 1)}}{2}\] here \[n = 6\]So, \[\dfrac{{6(6 - 1)}}{2} = 15\]Hence, the answer is option B which is 15 is correct. Note: Remember that an excited electron can return back directly and also in several steps. However, once it returns to its ground state, the energy released will be the same.
Since comments caused certain level of confusion, I guess I'll try to provide a further illustration. You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$. Electron doesn't get stuck forever on any of the levels with $n > 1$. Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms. In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$. The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this. I put together a rough drawing in Inkscape to illustrate all possible transitions*:  I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots). To determine $N$, you need to sum the states, as Soumik Das rightfully commented: $$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$ For $n = 6$: $$N = \frac{6(6-1)}{2} = 15$$ Obviously the same result is obtained by taking the sum directly. * Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$. |
What is the maximum number of spectral lines produced when an electron jumps from n 7 to n 1 is
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