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### Uncertainty relation results in rejection of semi classic Bohr model for hydrogen atom

Question
Show that the uncertainty relation forces us to reject the semi classical Bohr model for the hydrogen atom
Solution
In Bohr atom model we deal with the electron as a classical particle. The allowed orbits are defined by the quantization rules:
The radius r of a circular orbit and the momentum $p = mv$ of the rotating electron must satisfy $pr = n\hbar (n = 1,2,3,....)$. To consider an electron’s model in classical terms , the uncertainties in its position and momentum must be negligible when compared to  $r$ and $p$. In other words $\Delta x \ll r$ and $\Delta p \ll p$
This implies
$\frac{{\Delta x}}{r}\frac{{\Delta p}}{p} \ll 1$                     (1)
On the other hand , the uncertainty relation imposes
$\frac{{\Delta x}}{r}\frac{{\Delta p}}{p} \ge \frac{\hbar }{{rp}} \Rightarrow \frac{{\Delta x\Delta p}}{{rp}} \ge \frac{1}{n}$                           (2)
So, equation 1 is incompatible with 2 , unless $n \gg 1$

### Bohr hydrogen atom approaches classical condition

Question
prove that the Bohr hydrogen atom approaches classical conditions when n becomes very large and small quantum jumps are involved.
Solution
To prove this let us compute the frequency of a photon that is emitted in the transition between the adjacent state $n_{k}=n$ and $n_{i}=n-1$ when $n\gg 1$.
we define Rydberg’s constant as
$R=\frac{2\pi ^{2}me^{4}}{h^{3}c}$
So,
${E_k} = \frac{{ch}}{{n_k^2}}R$
and
${E_i} = \frac{{ch}}{{n_i^2}}R$
Therefore the frequency of the emitted photon is
$\nu = \frac{{n_k^2 - n_i^2}}{{n_k^2n_i^2}}cR = \frac{{\left( {{n_k} + {n_i}} \right)\left( {{n_k} - {n_i}} \right)}}{{n_k^2n_i^2}}cR$
${{n_k} - {n_i} = 1}$ , so for $n\gg 1$ we have
${n_k} + {n_i} \cong 2n$ and $n_k^2n_i^2 \cong {n^4}$
Therefore ,
$\nu = \frac{{2cR}}{{{n^3}}}$
According to classical theory of electromagnetism , a rotating charge with a frequency $f$ will emit a radiation of frequency $f$. On the other hand , using the Bohr hydrogen model , the orbital frequency of the electron around the nucleus is
${f_n} = \frac{{{\nu _n}}}{{2\pi {r_n}}} = \frac{{4{\pi ^2}m{e^4}}}{{{n^3}{h^3}}}$
or
${f_n} = \frac{{2cR}}{{{n^3}}}$, which is identical to $\nu$

### Parabolic motion (motion in two dimension)

Question
A stone is thrown from ground level over horizontal ground. It just clears three walls, the successive distances between them being r and 2r. The inner wall is 15/7 times as high as the outer walls which are equal in height. The total horizontal range is nr, where n is an integer. Find n.
Solution
Let us just assume that both the outer walls are equal in height say $h$ and they are at equal distance $x$ from the end points of the parabolic trajectory as can be shown below in the figure.

Now equation of the parabola is
$y = bx - c{x^2}$                                       (1)
$y = 0$ at $x = nr = R$
where $R$ is the range of the parabola.
Putting these values in equation (1) we get
$b = cnr$                                                       (2)
Now the range $R$ of the parabola is
$R = a + r + 2r + a = nr$
This gives
$a = \left( {n - 3} \right)\frac{r}{2}$               (3)
The trajectory of the stone passes through the top of the three walls whose coordinates are
$\left( {a,h} \right),\left( {a + r,\frac{{15}}{7}h} \right),\left( {a + 3r,h} \right)$
Using these co-ordinates in equation 1 we get
$h = ab - c{a^2}$                                           (4)
$\frac{{15}}{7}h = b(a + r) - c{(a + r)^2}$                              (5)
$h = b(a + 3r) - c{(a + 3r)^2}$                                      (6)
After combining (2), (3), (4), (5) and (6) and solving them we get n = 4.