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

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?

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?

The wavelength of radiation emitted is λ 0 when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be

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