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The orbital radii of a hydrogen - like atom is given by the equation

$$r_{n}=\frac{n^{2} \hbar^{2}}{Z m_{e} k_{e} e^{2}}$$

What is the radius of the first Bohr orbit in (a) $\mathrm{He}^{+},(\mathrm{b}) \mathrm{Li}^{2+}$

and $(\mathrm{c}) \mathrm{Be}^{3+} ?$

a) 0.0265 $\mathrm{mm}$

b) 0.0176 $\mathrm{nm}$

c) 0.0132 $\mathrm{nm}$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

University of Winnipeg

In this exercise, we have the information that the raid I of the orbits of hydrogen like atoms are given by and square times H bar square Invited by Z times the mass of the electron times of comes constant times The square where Z is the atomic number off the atom and in questions, A, B and C would have to find the rate I of the ground states of the helium atom Vienna ionized helium Adam the doubly ionized lithium Adam. So for the Navy are nice helium for a question, be a doubly ionized lithium. And for Russian see the three times ionized barely barely an Adam. Um So what I'm gonna do, first of all, is to request recast this equation for the raid. I a little more friendly. Ah, some I'm gonna remind you. First of all that, the more radius A zero I'm gonna write it in Blue is defined as H bar square over em Que he squared. Okay. And this is a constant in the very useful number. This is 5.29 times 10 to the minus 11 meters. So I noticed that our equation for our end can be more friendly Recast as ah, in, I swear, a zero divided by zero. Okay, so now, uh, you know, off our cases in the three questions A, B and C, we have to find the energy for the ground state. So an equals one for that reason are one equals a zero divided by Z. So we need to do is to soups it to the respective z's off these three atoms into the equation for our walk. So let's do it for the helium atom. We have an art one. It's equal to a zero ver two because the atomic number off the Islamists, too. So that's 51 29 times 10 to the minus 11 meters, divided by two. Okay, um, me for a question. And this is equal to 2.645 times 10 to the miners. 11th Peters in question be we have that for the delivery. Um, the atomic number of the lithium is three. So this is 5.29 10th into the minus 11 meters, divided by three. And this is equal to 1.763 times 10 to the minus 11th meters and finally for question. see you have it are one is equal to a zero divided by four since the atomic number of the very lien is four. So this is five points 29 times 10 to the minus 11 meters, divided by four. So this is one point 323 times sent to the minors 11 meters.

Universidade de Sao Paulo