Determine the wavelength of the emitted photons when the electron

Aedan Tyler

Aedan Tyler

Answered question

2022-05-20

Determine the wavelength of the emitted photons when the electron jumps from n= 4 state to n= 2 state for a Hydrogen atom. What are its color and the name of the spectral series?

Answer & Explanation

Superina0xb4i

Superina0xb4i

Beginner2022-05-21Added 17 answers

For hydrogen atomic number is 1.
Applying the formula,
z = 1 ,   n f = 2 ,   n i = 4 1 λ = ( 1.09677 × 10 7   m 1 ) ( 1 ) 2 [ 1 ( 2 ) 2 1 ( 4 ) 2 ] 1 λ = ( 1.09677 × 10 7   m 1 ) ( 1 ) 2 [ 4 1 16 ] λ = 16 3 × ( 1.09677 × 10 7 m 1 ) = 4.86 × 10 7 m λ = 486   n m
The wavelength of the emitted photon corresponds to visible range.
The blue color range in visible spectra is 450 nm -495 nm, hence the color is blue and the series is Balmer series.

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