Hydrogen atoms are excited by a laser to the n = 4 state and then allowed...

Question

Hydrogen atoms are excited by a laser to the n = 4 state and then allowed to emit. What is the maximum number of distinct emission spectral lines (lines of different wavelengths) that can be observed from this system?

A, 1

B, 2

C, 3

D, 4

E, 5

F, 6

G, 7

H, 8

Calculate the wavelength of the 4 -> 2 transition.

Number

λ= ? m

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The process of transition is the process of moving an electron from the lower energy level to a higher level of energy or vice versa. Ground state is less energy-efficient than that of the excited energy level. Excited-state is characterized by a higher amount of a quantum number. When a transition occurs an electron with the hv quantity of energy gets released which is also known as the constant of planks.

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Mia Miller · Accepted Answer

An atom of hydrogen emits electromagnetic waves when the initial state is greater than the final state (n_i> n_f), so the changes of state that will emits an electromagnetic wave are:

Our system must count the emissions when the initial state is lower that n or i =4.

When n_i =3
When n_i =2
When n _i =1
- This is the lowest possible state for an atom. At this point, the atom can't emit energy.

The maximum number of distinct emission lines is 6. This applies to all atomic numbers as well as m. They are not listed in the problem's header.

The formula to calculate wavelength is calculated when the atom changes from one state of an atom to another.

Where

So, starting from the change in state 4 →2

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