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Chapter 19: Problem 51

In Bohr's model, the atomic radius of the first orbit is \(r_{0}\), then the radius of the third orbit is (A) \(\frac{r_{0}}{9}\) (B) \(r_{0}\) (C) \(9 r_{0}\) (D) \(3 r_{0}\)

Short Answer

Expert verified
The radius of the third orbit in Bohr's model is 9 times the radius of the first orbit, so the correct answer is (C) \(9r_0\).

Step by step solution

01

Recall the formula relating the radius of an orbit to the principal quantum number in Bohr's model

In Bohr's model of the hydrogen atom, the radius of an orbit with principal quantum number "n" can be expressed as: \(r_n = r_0 * n^2\), where \(r_0\) is the atomic radius of the first orbit, and "n" is the principal quantum number.
02

Calculate the radius of the third orbit

To find the third orbit radius, we need to apply the formula for the third orbit, with the principal quantum number "n" being 3. Thus, we have: \(r_3 = r_0 * (3^2)\).
03

Simplify the expression to obtain the radius of the third orbit

We now need to evaluate \(3^2\), which is equal to 9. So the formula becomes: \(r_3 = r_0 * 9\).
04

Choose the correct answer

Comparing our result with the options given in the problem, we find that the correct answer is (C) \(9r_0\). The radius of the third orbit in Bohr's model is equal to nine times the radius of the first orbit.

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Most popular questions from this chapter

When the voltage applied to an \(x\)-ray tube is increased from \(10 \mathrm{kV}\) to \(20 \mathrm{kV}\) the wavelength interval between the \(K_{\alpha}\) line and the short wave cut off of the continuous \(x\)-ray spectrum increases by a factor 3 . Find the atomic number of element of which the tube anti-cathode is made. (Rydberg's constant \(=10^{7} \mathrm{~m}^{-1}\) )

When a \(U^{2.38}\) nucleus originally at rest, decays by emitting an alpha particle having a speed \(u\), the recoil speed of the residual nucleus is (A) \(\frac{4 u}{238}\) (B) \(-\frac{4 u}{234}\) (C) \(\frac{4 u}{234}\) (D) \(-\frac{4 u}{238}\)

The radius of hydrogen atom in its ground state is \(5.3 \times 10^{-11} \mathrm{~m} .\) After collision with an electron it is found to have a radius of \(21.2 \times 10^{-11} \mathrm{~m}\). What is the principal quantum number \(n\) of the final state of the atom?

Two radioactive elements \(R\) and \(S\) disintegrate as \(R \longrightarrow P+\alpha ; \lambda_{R}=4.5 \times 10^{-3}\) years \(^{-1}\) \(S \longrightarrow Q+\beta ; \lambda_{S}=3 \times 10^{-3}\) years \(^{-1}\) Starting with number of atoms of \(R\) and \(S\) in the ratio of \(2: 1\), this ratio after the lapse of three half-lives of \(R\) will be (A) \(3: 2\) (B) \(1: 3\) (C) \(1: 1\) (D) \(2: 1\)

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