The Ground State Orbit of a W+73 Atom: Is It Bohr-rific?

[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.

 

[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

 

[Doc Al]

Some comments and hints.

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??)

Looks good. That's the total energy of the n-th level.

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

What's the speed of the electron in the ground state? How does it compare to c?

While you're at it, figure out the speed and radius of the ground state orbit, according to the Bohr model.