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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* = .

*hc* = 6.63 `x` 10^{- 34} *J* *s* `x` 3.0 `x` 10^{8} `x` `x` = 1243 *eV* *nm* .

= = = 0.62 *nm* .

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* = *kA*^{ 2} = 400 (0.1)^{2} = 2 *J* .

*f* = = = = 1.42 *Hz* .

*n* = = = 2.12 `x` 10^{33}.

*E* = (*n* + 1)*hf* - *nhf* = *hf* = 6.63 `x` 10^{- 34} 1.42 = 9.4 `x` 10^{- 34} *J* .

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* - = - .

= - *eV* = - 0.5 *eV* = 1.36 *eV* .

= = = 917.2 *nm* .

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*/ , we have

*E* = = *eV* = = 41,433 *eV* ,

*V* = = `x` = 41,433 = 41,433 *V* .

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?

**Solution: **

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

= = .

From this, we can find the speed required of the electron as
*v* = = = 7.28 `x` 10^{7}.

*E* = *mv*^{ 2} = `x` 9.11 `x` 10^{- 31} *kg* `x` ^{2} `x` = 15.1 *keV* .

Suppose visible light of wavelength
= 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

*x* *p* = .

*v* = = = 115.8.

10/9/1997