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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A quantum of electromagnetic radiation has an energy of 2.0 keV. What is the associated wavelength?

We will use Planck's relation

E = hf = $\displaystyle{\frac{hc}{\lambda}}$.

A convenient expression for the constant hc in this and similar problems is

hc = 6.63 x 10- 34 J $\displaystyle\cdot$ s x 3.0 x 108$\displaystyle{\frac{\ {\:\rm m}}{\ {\:\rm s}}}$ x $\displaystyle{\frac{1\ {\:\rm eV}}{1.6\times 10^{-19}\ {\:\rm J}}}$ x $\displaystyle{\frac{1\ {\:\rm nm}}{1.0\times 10^{-9} \ {\:\rm m}}}$ = 1243 eV $\displaystyle\cdot$ nm .

With this, we find

$\displaystyle\lambda$ = $\displaystyle{\frac{hc}{E}}$ = $\displaystyle{\frac{1243 \ {\:\rm eV}\cdot\ {\:\rm nm}}{2000 \ {\:\rm eV}}}$ = 0.62 nm .

Thus, the associated wavelength is 0.62 nm.

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

A 5 kg mass attached to a spring of spring constant k = 400 N/m is undergoing simple harmonic motion on a flat frictionless surface with an amplitude of 10 cm. If we assume the energy levels are quantized according to the Planck relation E = nhf , what is the corresponding quantum number n ?

We first find the total energy associated with the system as

E = $\displaystyle{\textstyle\frac{1}{2}}$kA 2 = $\displaystyle{\textstyle\frac{1}{2}}$ $\displaystyle\cdot$ 400 $\displaystyle\cdot$ (0.1)2 = 2 J .

To equate this energy E to nhf , we need the frequency of oscillation f , which can be found as

f = $\displaystyle{\frac{\omega}{2\pi}}$ = $\displaystyle{\textstyle\frac{1}{2\pi}}$$\displaystyle\sqrt{\frac{k}{m}}$ = $\displaystyle{\textstyle\frac{1}{2\pi}}$$\displaystyle\sqrt{\frac{400}{5}}$ = 1.42 Hz .

We then find

n = $\displaystyle{\frac{E}{hf}}$ = $\displaystyle{\textstyle\frac{2}{6.63\times 10^{-34}\cdot 1.42}}$ = 2.12 x 1033.

In such regimes quantization effects can generally be ignored - this can be seen, for example, by calculating the energy difference $\Delta$E between the system in this state and the system in the next excited state, where n has increased by 1. We see

$\displaystyle\Delta$E = (n + 1)hf - nhf = hf = 6.63 x 10- 34 $\displaystyle\cdot$ 1.42 = 9.4 x 10- 34 J .

In other words, there is a difference of about 9.4 x 10- 34 J between the energy levels when the system is in this state. This is negligible, and so for all practical purposes the energy spectrum is continuous (i.e., classical).

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Red light of wavelength 670.0 nm produces photoelectrons from a certain metal which requires a stopping potential of 0.5 V to stop. What is the work function and threshold wavelength of the metal?

We shall use the Einstein relation

eV = hf - $\displaystyle\phi$ = $\displaystyle{\frac{hc}{\lambda}}$ - $\displaystyle\phi$.

With a stopping potential V = 0.5 V, we find

$\displaystyle\phi$ = $\displaystyle{\frac{hc}{\lambda}}$ - eV = $\displaystyle{\frac{1243 \ {\:\rm eV}\cdot\ {\:\rm nm}}{670 \ {\:\rm nm}}}$ - 0.5 eV = 1.36 eV .

This corresponds to a threshold wavelength of

$\displaystyle\lambda_{c}^{}$ = $\displaystyle{\frac{hc}{E}}$ = $\displaystyle{\frac{1243 \ {\:\rm eV}\cdot\ {\:\rm nm}}{1.36\ {\:\rm eV}}}$ = 917.2 nm .

Thus, the work function of the metal is 1.36 eV, with a corresponding threshold wavelength of 917.2 nm.

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

What minimum accelerating voltage would be required for an electron if it was to give up all of its energy in a collision with a target and produce an x-ray of wavelength 0.03 nm?

When an electron is accelerated through a potential difference V , it acquires an energy eV . If this energy went into producing a photon of energy E , with associated wavelength hc/$\lambda$ , we have

E = $\displaystyle{\frac{hc}{\lambda}}$ = eV = $\displaystyle{\frac{1243 \ {\:\rm eV}\cdot\ {\:\rm nm}}{0.03 \ {\:\rm nm}}}$ = 41,433 eV ,

from which we find

V = $\displaystyle{\frac{E}{e}}$ = $\displaystyle{\frac{41,433 \ {\:\rm eV}}{1.6\times 10^{19}\ {\:\rm C}}}$ x $\displaystyle{\frac{1.6\times 10^{-19}\ {\:\rm J}}{1\ {\:\rm eV}}}$ = 41,433$\displaystyle{\frac{\ {\:\rm J}}{\ {\:\rm C}}}$ = 41,433 V .

Thus, the accelerating potential required is 41,433 V.

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

The resolving power of a microscope is approximately equal to that of the wavelength of the light used. In an electron microscope, electrons are used as the ``light'' source. What kinetic energy of electrons is needed if the resolving power is required to be 10- 11 m, which would enable one to ``see'' an atom?

By the de Broglie hypothesis, the wavelength $\lambda$ of a particle is related to its momentum p by

$\displaystyle\lambda$ = $\displaystyle{\frac{h}{p}}$ = $\displaystyle{\frac{h}{mv}}$.

From this, we can find the speed required of the electron as

v = $\displaystyle{\frac{h}{m\lambda}}$ = $\displaystyle{\frac{6.63\times 10^{-34} \ {\:\rm J}\cdot\ {\:\rm s}}{9.11\times 10^{-31}\ {\:\rm kg}\cdot 10^{-11}\ {\:\rm m}}}$ = 7.28 x 107$\displaystyle{\frac{\ {\:\rm m}}{\ {\:\rm s}}}$.

This corresponds to a kinetic energy of

E = $\displaystyle{\textstyle\frac{1}{2}}$mv 2 = $\displaystyle{\textstyle\frac{1}{2}}$ x 9.11 x 10- 31 kg x $\displaystyle\left(7.28\times 10^7 \frac{\ {\:\rm m}}{\ {\:\rm s}}\right)$2 x $\displaystyle{\frac{1\ {\:\rm eV}}{1.6\times 10^{-19}\ {\:\rm J}}}$ = 15.1 keV .

The required kinetic energy of the electrons is then 15.1 keV.

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$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Suppose visible light of wavelength $\lambda$ = 5 x 10- 7 m is used to determine the position of an electron to within the wavelength of the light. What is the minimum uncertainty in the electron's speed?

We shall use the Heisenberg uncertainty relation

$\displaystyle\Delta$ x $\displaystyle\Delta$ p = $\displaystyle{\frac{h}{4\pi}}$.

With p = mv , and assuming negligible uncertainty in the mass of the electron, we find

$\displaystyle\Delta$ v = $\displaystyle{\frac{h}{4\pi m\Delta\,x}}$ = $\displaystyle{\frac{6.63\times 10^{-34} \ {\:\rm J}\cdot\ {\:\rm s}}{4\pi\cdot 9.11\times 10^{-31}\ {\:\rm kg}\cdot 5\times 10^{-7}\ {\:\rm m}}}$ = 115.8$\displaystyle{\frac{\ {\:\rm m}}{\ {\:\rm s}}}$.

Thus, the minimum uncertainty in the electron's speed is 115.8 m/s.

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Next: About this document ... Up: Quantum Physics Previous: The Uncertainty Principle