the hydrogen emission spectrum can be seen above. the lines are produced as electrons fall from an excited…

the hydrogen emission spectrum can be seen above. the lines are produced as electrons fall from an excited state back to their ground state. the different energy levels are represented by the quantum number n, where n = 1 is the 1st energy level. which electron would produce the violet color on the spectrum? a electron from n = 3 → n = 2 b electron from n = 4 → n = 2 c electron from n = 5 → n = 2 d electron from n = 6 → n = 2

the hydrogen emission spectrum can be seen above. the lines are produced as electrons fall from an excited state back to their ground state. the different energy levels are represented by the quantum number n, where n = 1 is the 1st energy level. which electron would produce the violet color on the spectrum? a electron from n = 3 → n = 2 b electron from n = 4 → n = 2 c electron from n = 5 → n = 2 d electron from n = 6 → n = 2

Answer

Explanation:

Step1: Understand the relationship between energy and color

In the hydrogen emission spectrum, the energy of the photon emitted when an electron falls from a higher energy level (n_{high}) to a lower energy level (n_{low}) is given by (E = h\nu), where (h) is Planck's constant and (\nu) is the frequency of the photon. Violet light has a higher frequency (and thus higher energy) compared to other colors in the visible spectrum. The energy of the photon emitted during an electron transition is also related to the energy levels by (E=E_{n_{high}}-E_{n_{low}}). The energy of an electron in a hydrogen atom is (E_n =-\frac{13.6}{n^{2}}\text{ eV}).

Step2: Calculate the energy difference for each transition

  • For (n = 3\rightarrow n = 2): (E_3=-\frac{13.6}{3^{2}}\text{ eV}\approx - 1.51\text{ eV}), (E_2=-\frac{13.6}{2^{2}}\text{ eV}=- 3.4\text{ eV}) (\Delta E=E_3 - E_2=(-1.51)-(-3.4)=1.89\text{ eV})
  • For (n = 4\rightarrow n = 2): (E_4=-\frac{13.6}{4^{2}}\text{ eV}=- 0.85\text{ eV}) (\Delta E=E_4 - E_2=(- 0.85)-(-3.4)=2.55\text{ eV})
  • For (n = 5\rightarrow n = 2): (E_5=-\frac{13.6}{5^{2}}\text{ eV}=- 0.544\text{ eV}) (\Delta E=E_5 - E_2=(-0.544)-(-3.4)=2.856\text{ eV})
  • For (n = 6\rightarrow n = 2): (E_6=-\frac{13.6}{6^{2}}\text{ eV}\approx - 0.38\text{ eV}) (\Delta E=E_6 - E_2=(-0.38)-(-3.4)=3.02\text{ eV})

Step3: Relate energy to color

Since violet light has a higher energy among the visible lines in the Balmer series ((n_{low}=2)), and among the given transitions, the transition (n = 6\rightarrow n = 2) has the highest energy difference.

Answer:

D. Electron from (n = 6\rightarrow n = 2)