Question #5745c

1 Answer
Feb 11, 2017

Answer:

#v_4/v_7=7/4=1.75#

Explanation:

From Rutherford-Bohr's atomic model of Hydrogen we know that energy levels or velocity of an electron is defined by an positive integer #n#, where #n=1,2,3,......oo#. This was later defined as principal quantum number.

Assuming that transition here implies movement of an electron from higher to lower value of #n#, an electron which can have maximum #x# transitions would be have #n=(x+1)#. As #n=1# is ground state.

Therefore, for an electron which can have maximum of #6# transitions, #n=7#.
Similarly, for an electron which can have maximum of #3# transitions, #n=4#.

One of Bohr's key hypotheses proposed was that the orbiting electron could exist only in certain special states called stationary states. In these states, the angular momentum of the electron #vecL# assumes only integer values of Planck's constant divided by #2pi#, so that no electromagnetic radiation was emitted.

For a circular orbit we have momentum of an electron #vecL=vecrxxvecp=m_evr# which can have values #(nh)/(2pi)#, where #n# has already been defined above.

From above and solving for velocity we have
#v=(nh)/(2pim_er)# ....(1)
Also recognizing that for a stable orbit Coulomb's force of attraction is equal and opposite to the centripetal force we get
#1/(4piepsilon_0)e^2/r^2=(m_ev^2)/r#
Using (1) we get
.#1/(4piepsilon_0)e^2/r^2=m_e/r((nh)/(2pim_er))^2#
Rearranging we get allowed radii as

#r_n=a_0n^2# ......(2)
where #a_0=(4piepsilon_0)/m_e((h)/(2pie))^2=0.0529" nm"#, Bohr's radius.

From (1) and (2) we get allowed velocities as
#v_n=h/(2pim_ea_0n)# .....(3)

The required ratio is

#v_4/v_7=7/4=1.75#