What series of lines are observed in the hydrogen spectrum?

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  • The hydrogen emission spectrum contains several spectral series, each named after its discoverer.

    • Lyman series (n' = 1): To ground state, visible only in ultraviolet region
    • Balmer series (n' = 2): Transitions to second level, appearing in visible region
    • Paschen series (n' = 3): Moved to third level, visible in the infrared region
    • Brackett series (n' = 4): Transitions to fourth level, appearing in the far infrared region
    • Pfund series (n' = 5): Transitions to fifth level, showing in the infrared region
    • Humphreys series (n' = 6): Transitions to sixth level, appearing in the infrared region

Similar Questions for you

S
Syed Aquib Ur Rahman

Hydrogen shows many spectral lines because of the following reasons. 

  • Its electron can occupy many levels (n = 1, 2, 3.). 

  • Each line in the hydrogen spectrum actually represents the transition from higher to lower energy levels. 

  • These lines are grouped by the final energy level (Lyman n'=1, Balmer n'=2, etc.) 

  • Higher energy levels are closer together. They tend to create more possible transitions.

S
Syed Aquib Ur Rahman

Hydrogen produces a line spectrum because electrons exist only in discrete, quantised energy levels. When electrons jump between these fixed energy states, they emit photons with specific energies (E = h? ). This creates distinct spectral lines instead of continuous wavelengths. Bohr's model explains this through quantised orbits. Classical physics, on the other hand, would predict a continuous spectrum.

A
alok kumar singh

Change in surface energy = work done

 

A
alok kumar singh

| Δ E 0 | = ( 1 3 6 { 1 1 4 } ) e V

|DE0| = –10.2

λ = 1 2 4 0 0 1 0 . 2 × 1 0 1 0 m

ρ = h λ = 6 . 6 3 × 1 0 3 4 × 1 0 . 2 1 2 4 0 0 × 1 0 1 0                  

? m v = h λ            

  1 . 8 × 1 0 2 7           

v = 6 . 6 3 × 1 0 . 2 × 1 0 3 4 1 2 4 0 0 × 1 0 1 0

v = 6 . 6 3 × 1 0 . 2 1 2 4 0 0 × 1 . 8 × 1 0 3            

= 6 . 6 3 × 1 0 2 1 2 4 × 1 . 8 = 3 . 0 2 ]

= 3 m/s

A
alok kumar singh

0 . 8 5 = 1 3 . 6 n 2  

n = 4

Number of transitions = 4 × 3 2 = 6  

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