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Chapter 38: Q. 58 (page 1116)

Starting from Equation 38.32, derive Equation 38.33

Short Answer

Expert verified

The Derive equation is $E_{n} = - \frac{1}{n^{2}} (\frac{1}{4 {πϵ}_{0}} \frac{e^{2}}{2 a_{B}})$

Step by step solution

01

Given information

In atomic physics, the Bohr radius is a physical constant. It discusses the most likely distance between a hydrogen atom's nucleus and electron in its ground state.

02

Calculations

We can begin with the expression (38.32) and the Bohr radius expression.

$E_{n} = \frac{1}{2} m (\frac{ħ^{2}}{m^{2} a_{B}^{2} n^{2}}) - \frac{e^{2}}{4 {πϵ}_{0} n^{2} a_{B}} a_{B} = \frac{4 {πϵ}_{0} ħ^{2}}{{me}^{2}} \frac{ħ^{2}}{{ma}_{B}} = \frac{e^{2}}{4 {πϵ}_{0}} (express) E_{n} = \frac{1}{2} \frac{e^{2}}{4 {πϵ}_{0} n^{2} a_{B}} - \frac{e^{2}}{4 {πϵ}_{0} n^{2} a_{B}} (substitute previous expression) E_{n} = - \frac{1}{n^{2}} (\frac{1}{4 {πϵ}_{0}} \frac{e^{2}}{2 a_{B}})$

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Most popular questions from this chapter

A photoelectric-effect experiment finds a stopping potential of $1.56 V$ when light of $200 nm$ is used to illuminate the cathode.

a. From what metal is the cathode made?

b. What is the stopping potential if the intensity of the light is doubled?

A proton emits a gamma-ray photon with energy 2.0 MeV in a quantum jump from n =2 to n= 1.

a. A negatively charged electroscope can be discharged by shining an ultraviolet light on it. How does this happen?

b. You might think that an ultraviolet light shining on an initially uncharged electroscope would cause the electroscope to become positively charged as photoelectrons are emitted. In fact, ultraviolet light has no noticeable effect on an uncharged electroscope. Why not?

I What is the radius of a hydrogen atom whose electron is bound by $0.378 eV$ ?

a. What quantum number of the hydrogen atom comes closest to giving a 100-nm-diameter electron orbit?

b. What are the electron’s speed and energy in this state?

See all solutions

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