# Question #5745c

Feb 11, 2017

${v}_{4} / {v}_{7} = \frac{7}{4} = 1.75$

#### 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 $n = 1 , 2 , 3 , \ldots \ldots \infty$. 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 $n = \left(x + 1\right)$. 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 $\vec{L}$ assumes only integer values of Planck's constant divided by $2 \pi$, so that no electromagnetic radiation was emitted.

For a circular orbit we have momentum of an electron $\vec{L} = \vec{r} \times \vec{p} = {m}_{e} v r$ which can have values $\frac{n h}{2 \pi}$, where $n$ has already been defined above.

From above and solving for velocity we have
$v = \frac{n h}{2 \pi {m}_{e} r}$ ....(1)
Also recognizing that for a stable orbit Coulomb's force of attraction is equal and opposite to the centripetal force we get
$\frac{1}{4 \pi {\epsilon}_{0}} {e}^{2} / {r}^{2} = \frac{{m}_{e} {v}^{2}}{r}$
Using (1) we get
.$\frac{1}{4 \pi {\epsilon}_{0}} {e}^{2} / {r}^{2} = {m}_{e} / r {\left(\frac{n h}{2 \pi {m}_{e} r}\right)}^{2}$
Rearranging we get allowed radii as

${r}_{n} = {a}_{0} {n}^{2}$ ......(2)
where ${a}_{0} = \frac{4 \pi {\epsilon}_{0}}{m} _ e {\left(\frac{h}{2 \pi e}\right)}^{2} = 0.0529 \text{ nm}$, Bohr's radius.

From (1) and (2) we get allowed velocities as
${v}_{n} = \frac{h}{2 \pi {m}_{e} {a}_{0} n}$ .....(3)

The required ratio is

${v}_{4} / {v}_{7} = \frac{7}{4} = 1.75$