What is the wavelength of a photon emitted during the electron transition from n 5 to n 3 state in hydrogen atom?

What are the frequency and wavelength of a photon emitted during a transition from n=5 state to the n=2 state in the hydrogen atom ?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Suggest Corrections

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

Text Solution

Solution : since `n_(1) = 5 and n_(1) = 2 `, this transition gives rise to a spectral line in the visible region of the balmer series. From equation `DeltaE=2.18xx10^(-18)J[1/5^(2)-1/2^(2)]` ` 4.58xx 10^(-19)J` it is an emission energy the frequency of the photon ( taking energy in terms of magnitude ) is given by `v=(DeltaE)/h` `( 4.58xx10^(-19)J)/(6.626xx10^(-34) Js)` `6.91xx10^(14)Hz` ` lambda=c/v= (3.0xx10^(8)m s^(-1))/(6.91xx10^(14)Hz)=434 nm`

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"#
#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"#