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Verify equations $r_n = \dfrac{n^2}{Z}a_B$ and $a_B = \dfrac{h^2}{4\pi^2m_ekq_e^2} = 0.529 \times 10^{-10}\textrm{ m}$ using the approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization.
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OpenStax College Physics Solution, Chapter 30, Problem 23 (Problems & Exercises) (4:53)

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This is College Physics Answers with Shaun Dychko. We have this formula telling us that the radius of a hydrogen-like atom which is to say either a hydrogen atom or an ion that has only 1 electron is the principal quantum number squared divided by the atomic number which is the number of protons multiplied by the Bohr radius. And we are going to verify that that's true and then we'll also calculate what the Bohr radius is by plugging in for each of the constants here. So we begin by saying that the electrostatic force that's holding the electron around the positive nucleus equals the centripetal force and this isn't strictly true that the electron is a ball that's moving in a circle around some nucleus like the Earth around the Sun. But for some reason, this mathematically it works out that way even though conceptually the electron is more of a a way that's existing in a cloud of probability that it's somewhere around. In any case, we can say this for a hydrogen-like atom and it will work out to be correct. So we have coulombs constant times the charge of the nucleus which is the number of protons multiplied by the elemantary charge and then we multiply it by the single electron that's orbiting around and so that's why we get <i>Z q e</i> squared and then we divide by the radius of the atom squared. And this is gonna equal the centripetal force which is mass times speed squared over the orbital radius. Now we are told to make a substitution for based on the quantization of angular momentum. So angular momentum based on equation [30.20] is going to be quantized as the principal quantum number multiplied by Planck's constant over 2<i>π</i> where <i>n</i> is the number 1, 2, 3 and so on some natural number. And so we solve this for <i>v</i> and so it is by dividing both sides by mass of the electron times the radius. And then we have <i>v</i> is <i>nh</i> over mass of the electron times the radius of the atom or ion times 2<i>π</i>. So we substitute that for <i>v</i> up here. And well there's a whole bunch of letters here and our job is to solve it for <i>r n</i>. So after we square this bracket, squaring the <i>v</i> in other words, oh and another thing I did here is this is <i>r n</i> squared and this is <i>r n</i> multiplied both sides by <i>r n</i>. So that makes this <i>r n</i> disappear and this becomes <i>r n</i> to the power of 1 here. And so we have <i>kZq e</i> squared over <i>r n</i> equals <i>n</i> squared <i>h</i> squared over <i>m e</i> to the power of 1 because this <i>m e</i> in the denominator gets squared multiplying by this <i>m e</i> makes it <i>m e</i> to the power of 1 in the bottom times <i>r n</i> squared times 4<i>π</i> squared. And then multiply both sides by <i>r n</i> <i>r n</i> squared in fact and this cancels and over here you get <i>r n</i> to the power of 1 and so that's what's here. And then we'll divide both sides by all of this and we end up with this line here. So <i>r n</i> is <i>n</i> squared <i>h</i> squared in the numerator divided by <i>Z</i> and <i>k</i> and <i>q e</i> squared and <i>m e</i> and 4 and <i>π</i> squared. And all of this is the formula for the Bohr radius and so we have verified that the radius of the atom is going to be the principal quantum number squared over the atomic number times Bohr radius. And we arrived at this by beginning with this equating of the coulomb electrostatic force with the centripetal force. Then we substitute for Planck's constant to calculate the Bohr radius; Planck's constant squared divided by mass of the electron times 4<i>π</i> squared times coulomb's constant times the elementary charge squared and we get 0.529 times 10 to the minus 10 meters and so we have shown that as well.