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

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.

• How do you get these values? You assume that r is the same for both values of n?
– nasu
Commented Aug 24, 2016 at 19:10
• @nasu , i assume r(n=1) as the Bohr radius(a) and for different values of n we have r = a*n^2 Commented Aug 25, 2016 at 1:53
• The problem is in your assumption that the energy of the bound electron will decrease as the wavelength increases. In this model the electron has both kinetic and potential energy (and the total energy is negative). If you do the calculations you see that the energy decreases with n and wavelength increases with n.
– nasu
Commented Aug 25, 2016 at 22:07
• well, the energy tends to zero as n tends to infinity. Energy is indeed negative(bound system) but it increses with n...for example, for H- atom; when n=1, E=-13.6 ev; n= 2, E = -3.4 ev and so on....so clearly E increases with n but from the postulate , De-Broglie wavelength increases with n....i found out that the answer is quite complicated than we can imagine, i will try and post it here. Commented Aug 27, 2016 at 2:00

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).

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.