Bohr / De Broglie postulate (what does $n λ= 2 π r$ imply)

Question

From the Bohr/De Broglie postulate we have n λ = 2πr where λ is the De-Broglie wavelength , r is the radius corresponding to n and n is the quantum number.

1. An electron in the state n=2 has more energy than that at n=1
2. That implies that the De- Broglie wavelength associated with the electron should also decrease ?

From the postulate..it is the other way i.e. the wavelength increases as the electron gains energy. How is this possible?.( I had assumed that wavelength decreases with energy)

if we calculate the De-Broglie wavelengths from the postulate:

for n=1 ; λ = 33 * 10^-11 m

for n=2 ; λ = 66 * 10^-11 m

does this mean that as the energy of the electron increases the corresponding De-Broglie wavelength increases?! may be i am missing something very basic here.

Answer 1

Well, yes. You are missing something basic here. The Bohr/de Broglie postulate $$n\lambda=2\pi r \tag{1}$$ alone is not enough to derive anything, because this one equation contains two unknowns ($\lambda$ and $r$). So you need more equations.

You get another equation from the de Broglie relation, which unfortunately introduces one more unknown ($v$): $$\lambda=\frac{h}{mv} \tag{2}$$

You get a third equation from the equality of centripetal force and Coulomb force: $$\frac{mv^2}{r}=\frac{Ze^2}{4\pi\epsilon_0 r^2} \tag{3}$$

And now you have three equations (1, 2, 3) for three unknowns ($\lambda$, $r$, $v$).

Doing the math (and I recommend you do this on your own sheet of paper) you can resolve these equations for the unknowns: $$\begin{align} \lambda &= \frac{2\epsilon_0 h^2}{Ze^2m}n &= 33\cdot 10^{-11}\text{ m}\cdot n \\ r &= \frac{\epsilon_0h^2}{\pi Ze^2m}n^2 &= 5.3\cdot 10^{-11}\text{ m}\cdot n^2 \\ v &= \frac{Ze^2}{2\epsilon_0h}\frac{1}{n} \end{align}$$ From these you can also calculate the total energy (kinetic + potential energy).

Answer 2

Well if you go with Bohr's model of atom then, since $mv^2/2 = K.E.= KZe/(2an^2)$

$= 2.18 *10^{-18} *Z^2/n^2$.

It means the total kinetic energy decreases with increase in n. This also means that velcity, $v$ of electron also decreases with n.

Now look at de Broglie's equation- $\lambda = h/m_ev$

Since velocity decreases with increase in $n$, with increase in $n$ there is increase in $\lambda$. The total energy doesn't tell anything about velocity of electron. It is the kinetic energy that tells about velocity.