Question

Question #5745c

Answer 1

`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`