Problems

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

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

Solution:
We will use Planck's relation

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

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

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

With this, we find

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

Thus, the associated wavelength is 0.62 nm.

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

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 ?

Solution:
We first find the total energy associated with the system as

E = $${\textstyle\frac{1}{2}}$$kA ² = $${\textstyle\frac{1}{2}}$$ $$\cdot$$ 400 $$\cdot$$ (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 = $${\frac{\omega}{2\pi}}$$ = $${\textstyle\frac{1}{2\pi}}$$$$\sqrt{\frac{k}{m}}$$ = $${\textstyle\frac{1}{2\pi}}$$$$\sqrt{\frac{400}{5}}$$ = 1.42 Hz .

We then find

n = $${\frac{E}{hf}}$$ = $${\textstyle\frac{2}{6.63\times 10^{-34}\cdot 1.42}}$$ = 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

$$\Delta$$E = (n + 1)hf - nhf = hf = 6.63 x 10^{- 34} $$\cdot$$ 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).

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

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?

Solution:
We shall use the Einstein relation

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

With a stopping potential V = 0.5 V, we find

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

This corresponds to a threshold wavelength of

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

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

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

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?

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

from which we find

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

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

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

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<sup>- 11</sup> m, which would enable one to ``see'' an atom?

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

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

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

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

This corresponds to a kinetic energy of

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

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

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

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?

Solution:
We shall use the Heisenberg uncertainty relation

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

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

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

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