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; #n_("initial")# - the initial energy level - in your case equal to 5. 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"# 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 Open in App 0Calculate 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. Open in App 0Calculate 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? Open in App Suggest Corrections Open in App Suggest Corrections 0 |