What is the wavenumber for longest wavelength transition in Lyman series?

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

We know Energy of the incident radiation = Work function + KE of photoelectron

Work function = Energy of the incident

radiation - KE of photoelectron ..(1).Now energy of the incident radiation (E) = hv

..(2)

Substituting the values in eq. (2), we have

Energy of incident radiation

The potential applied gives the kinetic energy to the electron. Hence, the kinetic energy of the electron = 4.4 eV.Substituting the values in eq. (1), we have Work function = 4.83 eV - 0.35 eV = 4.48 eV

Text Solution

Solution : For Balmer series, `n_(1)=2`. Hence, `bar(v)=R(1/2^(2)-1/n_(2)^(2))` `bar(v)=1/lambda`. For `lambda` to be longest (maximum), `bar(v)` should be minimum. This can be so when `n_(2)` is minimum, i.e., `n_(2)=3`. Hence, `bar(v)=(1.097xx10^(7) m^(-1)) (1/2^(2)-1/3^(2))=1.097xx10^(7)xx5/36 m^(-1) =1.523xx10^(6) m^(-1)`

Text Solution

Solution : For Balmer series ` n_1 =2` If this line possesses longest wavelength ( i.e., lowest energy ) then ` n_2 =3` ` overline (v) = 1/ (lambda) = 109677 [1/2^2 - 1/3^2]` ` = 1. 523 xx 10^4 cm^(-1) = 1. 523 xx 10^6 m^(-1)`.

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

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