  Question:

What is the wave length of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2? Please explain the answer in brief

Posted by: Jeetendra B. on 26.09.2017 Sonam M. from Pune
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According to Balmer's Formula,

Wave no. = RH [1/n1^2 – 1/n2^2]

where RH Rydberg constant for hydrogen

Here n1 = 2, n2 = 4, RH = 109678

Putting these values in the equation we get

Wave no. = 109678 (1/2^2 - 1/4^2) = 109678  (1/4-1/16) = 109678 * 12/64 = 109678 * 3/16

Also λ = 1/ wave number

Therefore λ = 16 / 109678 x 3 = 486 nm Ritesh J. from Bangalore
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According to Balmer formula,

Wave no = RH [(1/n1)^2 – (1/n2)^2]

Here n1 = 2 ,       n2 = 4,     RH = 109678

Putting these values in the equation we get

Wave number =109678 * [((1/2)^2) - ((¼) ^ 2))] = 109678 x (3/16)

Also λ = 1/ wave number

Therefore λ = 16 / 109678 x 3 = 486 nm Rishav S. from Mumbai
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E= E0{(1/(n1)^2)-(1/(n2)^2)} eV

E= 13.6{1/4- 1/16} eV

E= 13.6x3/16 eV

E= 2.55 eV

hc/λ= 2.55 eV

λ= 4.8 x 10-7 m

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