How many transitions are possible when an electron jumps from N 5 to N 1 shell in a hydrogen atom?

The energy transition will be equal to #1.55 * 10^(-19)"J"#.

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

#1/(lamda) = R * (1/n_("final")^(2) - 1/n_("initial")^(2))#, where

#lamda# - the wavelength of the emitted photon;
#R# - Rydberg's constant - #1.0974 * 10^(7)"m"^(-1)#;
#n_("final")# - the final energy level - in your case equal to 3;
#n_("initial")# - the initial energy level - in your case equal to 5.

So, you've got all you need to solve for #lamda#, so

#1/(lamda) = 1.0974 * 10^(7)"m"^(-1) * (1/3^2 - 1/5^2)#

#1/(lamda) = 0.07804 * 10^(7)"m"^(-1) => lamda = 1.28 * 10^(-6)"m"#

Since #E = (hc)/(lamda)#, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by #h * c#, where

#h# - Planck's constant - #6.626 * 10^(-34)"J" * "s"#
#c# - the speed of light - #"299,792,458 m/s"#

So, the transition energy for your particular transition (which is part of the Paschen Series) is

#E = (6.626 * 10^(-34)"J" * cancel("s") * "299,792,458" cancel("m/s"))/(1.28 * 10^(-6)cancel("m"))#

#E = 1.55 * 10^(-19)"J"#

Answer

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Hint: Hydrogen atom consists of only one electron. But the emission spectrum consists of several lines. The various possibilities by which an electron is excited to a higher energy level s known electronic transitions. The series of spectral lines in the emission spectrum can be classified as five series. K shell has the principal quantum number \[n = 1\] and the O shell has the principal quantum number \[n = 5\]. It symbolizes that an electron jumps from fifth to first energy level.

Complete answer:

In the question, it is given that an electron jumps from fifth to first energy level. Thus, a total of ten distinct lines are observed in the hydrogen spectrum as four Lyman series, three Balmer series, two Paschen series, and one bracket series.The possible transitions for an electron jump from O shell to K shell can be written as,\[O \to N \to M \to L \to K\]\[O \to N \to L \to K\]\[O \to N \to K\]\[O \to M \to K\]\[O \to L \to K\]\[O \to K\] Here, some of them may consider \[O \to M \to L \to K\] as one of the possibilities of spectral lines. But in question, a maximum transition is needed. \[M \to L\] Is a small transition from third to second energy levels. So, it is not considered.Thus, there are six minimum H-atoms are required to observe a maximum number of distinct spectral lines where an e jumps from O shell to K shell

Hence, option C is the correct option..

Note: Lyman series are produced when an electron jumps from any of the higher levels to the first energy level. Balmer series are produced when an electron jumps from any of the higher levels to the second energy level. Paschen series is produced when an electron jumps from any of the higher levels to the third energy level. Bracket series are produced when an electron jumps from any of the higher levels to the fourth energy level. Pfund series are produced when an electron jumps from any of the higher levels to the fifth energy level.

The energy transition will be equal to #1.55 * 10^(-19)"J"#.

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

#1/(lamda) = R * (1/n_("final")^(2) - 1/n_("initial")^(2))#, where

#lamda# - the wavelength of the emitted photon; #R# - Rydberg's constant - #1.0974 * 10^(7)"m"^(-1)#;

#n_("final")# - the final energy level - in your case equal to 3;

#n_("initial")# - the initial energy level - in your case equal to 5.

So, you've got all you need to solve for #lamda#, so

#1/(lamda) = 1.0974 * 10^(7)"m"^(-1) * (1/3^2 - 1/5^2)#

#1/(lamda) = 0.07804 * 10^(7)"m"^(-1) => lamda = 1.28 * 10^(-6)"m"#

Since #E = (hc)/(lamda)#, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by #h * c#, where

#h# - Planck's constant - #6.626 * 10^(-34)"J" * "s"#
#c# - the speed of light - #"299,792,458 m/s"#

So, the transition energy for your particular transition (which is part of the Paschen Series) is

#E = (6.626 * 10^(-34)"J" * cancel("s") * "299,792,458" cancel("m/s"))/(1.28 * 10^(-6)cancel("m"))#

#E = 1.55 * 10^(-19)"J"#

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 $N$.

How many spectral line will be observed when an electron jumps from n=5 to n=1 in the visible line spectra of H spectrum?

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How many spectral line will be observed when an electron jumps from n=5 to n=1 in the visible line spectra of H spectrum?

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Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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