What is the wavelength of the radiation emitted when electron in hydrogen atom jumps from infinity to 2?

The energy transition will be equal to #1.55 * 10^(-19)"J"#.

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

#1/(lamda) = R * (1/n_("final")^(2) - 1/n_("initial")^(2))#, where

#lamda# - the wavelength of the emitted photon; #R# - Rydberg's constant - #1.0974 * 10^(7)"m"^(-1)#;

#n_("final")# - the final energy level - in your case equal to 3;

So, you've got all you need to solve for #lamda#, so

#1/(lamda) = 1.0974 * 10^(7)"m"^(-1) * (1/3^2 - 1/5^2)#

#1/(lamda) = 0.07804 * 10^(7)"m"^(-1) => lamda = 1.28 * 10^(-6)"m"#

Since #E = (hc)/(lamda)#, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by #h * c#, where

#h# - Planck's constant - #6.626 * 10^(-34)"J" * "s"#
#c# - the speed of light - #"299,792,458 m/s"#

So, the transition energy for your particular transition (which is part of the Paschen Series) is

#E = (6.626 * 10^(-34)"J" * cancel("s") * "299,792,458" cancel("m/s"))/(1.28 * 10^(-6)cancel("m"))#

#E = 1.55 * 10^(-19)"J"#

When an electron jumps from a radius n=∞ to n=2 of hydrogen atom then it radiates energy of the wavelength nm . If required, use hc = 1242 eV . nm

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Calculate the wavelength of radiation emitted when electron in a hydrogen atom jumps from n = `oo` to n = 1.

The electron in the hydrogen atom jumps from n2 = `oo`to n1 = 1. Now, we know that

`:.1/lambda= R(1/(n_1^2)-1/(n_2^2))`

`:.1/lambda=1.097xx10^7(1/1-1/oo)=1.097xx10^7`

`:.1/(1.097xx10^7)=9.116xx10^-8" " m= 911.6xx10^(-10)m`

 ∴ λ = 911.6 Å

Concept: Dual Nature of Radiation

  Is there an error in this question or solution?

Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.

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Calculate the wavelength of radiation emitted when electron in a hydrogen atom jumps from n = `oo` to n = 1.

The electron in the hydrogen atom jumps from n2 = `oo`to n1 = 1. Now, we know that

`:.1/lambda= R(1/(n_1^2)-1/(n_2^2))`

`:.1/lambda=1.097xx10^7(1/1-1/oo)=1.097xx10^7`

`:.1/(1.097xx10^7)=9.116xx10^-8" " m= 911.6xx10^(-10)m`

 ∴ λ = 911.6 Å

Concept: Dual Nature of Radiation

  Is there an error in this question or solution?

Calculate the wavelength of radiation emitted when electron in a hydrogen atom jumps from n = `oo` to n = 1.

The electron in the hydrogen atom jumps from n2 = `oo`to n1 = 1. Now, we know that

`:.1/lambda= R(1/(n_1^2)-1/(n_2^2))`

`:.1/lambda=1.097xx10^7(1/1-1/oo)=1.097xx10^7`

`:.1/(1.097xx10^7)=9.116xx10^-8" " m= 911.6xx10^(-10)m`

 ∴ λ = 911.6 Å

Concept: Dual Nature of Radiation

  Is there an error in this question or solution?

When an electron jumps from a radius n=∞ to n=2 of hydrogen atom then it radiates energy of the wavelength nm . If required, use hc = 1242 eV . nm

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The wavelength of the radiation emitted when a hydrogen atom electron falls from infinity to 2 nd excited state would be: Rydberg constant =1.097 × .107 m 1A. 406 nmB. 192 nmC. 168 nmD. 821 nm

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