Question

In Lyman series, one wavelength in the ultraviolet part of the hydrogen emission spectrum has wavelength 97.2 nm.
a) What is the initial state of the transition that results in this wavelength
being emitted? b) What is the radius of the orbit for the initial state of the atom? c) What is the energy of the initial and final state? d) What would be the velocity of an electron of hydrogen atom in an orbit
for n = 3 according to the Bohr model?

Answer 1

Rydberg formula for Hydrogen spectrum is given by

..................(1)

where h = 6.626 10^-34 Js is plancks constant, c is speed of light, is the wavelength when transition take place from n_i to n_f levels of energy states, m is mass of electron, e is charge of electrn , emissivity of free space

The term corresponds to ground state energy level of hydrogen atom .

Hence we have ,

............(2)

For Lyman series , final state n_f = 1

Hence using equation (2) and sunstituting other values , we get wavelength of a line in lyman series from eqn.(1) as

.........................(3)

to get a wavelenth of 97.2 nm , from above equation , we get initial level n_i = 4

----------------------------

Radius of orbit of n^th energy level is obtained from following equation

Hence by substituting n = 4 and after simplification we get radius of orbit that corresponds to energy level with n= 4 as

------------------------------------------------------------

n^th state Energy level E_n of Hydrogen atom is given by

Hence for the state n = 4, we get

Ground state energy level E₁ = -13.6 eV

----------------------------------------------------------------

velocity of an electron of hydrogen atom in an orbit that has quantum number n is given by

Hence fr n = 3 , we have

-------------------------------------------------------------------