# Find the orbital speed

1. Nov 14, 2004

### stunner5000pt

The nucleus of Tungsten conssits of 74protons and 110 neutrons. SUppose all of the electrons were stripped off, but one, leaving a W+73 atom.
Using the Bohr idea of the atom (not anything else!)
How much energy nrequired to remove this last electron from orbit??

The energy of the elctron in this state is

$$E = -Z^2 \frac{m e^4}{8 \epsilon_{0}^2 h^2 n}$$

and the energy to strip this electron from orbit is equal to this energy (yes??)

b) is the ususal assumption for the Bohr atom being nonrelaiivistic good for this atom?

i woul think yes because this atom only consists of two bodies and is very much like a much higher charged hydrogen atom

c)What is the wavelength of hte ecetron is this orbit?

first i must find the orbital speed but $$v = \frac{Ze^2}{2 \epsilon_{0} h} \frac{1}{n}$$

and then use $$\lambda = \frac{h}{m_{e} v}$$

d)Radius of the tungsten atom is 10^-13m Does the idea of a Bohr atom apply for the ground state of the W+73 atom?

I owuld think yes because a bohr atom is a pointcharge oribiting a heavy nucleus.

Last edited: Nov 14, 2004
2. Nov 14, 2004

### stunner5000pt

i have a correction for the B part

since the radius is given to be 10^-13 metres

$$v = \sqrt{\frac{Ze^2}{4 \pi \epsilon_{0} r}$$ and thereafter the velocity can be found

3. Nov 15, 2004