**PHY1161C
**

December 6, 2000

**Possibly useful information:**

c = 3.00 x 10^{8} m/s h
= 6.626 x 10^{-34} Js = 4.14 x 10^{-15} eV s

_{
}_{
}_{
}_{}

E = m c^{2 }E
= h f

KE_{max} = h f - W_{o}
= h/p = h / mv** _{
}**(x)
(p

R = 1.097 x 10

,

1 u = 931 MeV/c^{2}

_{
u
Mev/c2
kg
electron
0.00054858
0.511
9.1094 x 10 - 31
proton
1.007276
938.27
1.67262 x 10 - 27
neutron
1.008665
939.57
1.67493 x 10 - 27
hydrogen
1.007 825
}

_{}--->_{}+_{}Q = ( m_{P}- m_{D}- m_{a}) c^{2 }_{}--- >_{}+ b^{-}+ Q = ( m_{P}- m_{D}) c^{2}

--- > + b^{+}+ n Q = ( m_{P}- m_{D}- 2 m_{b}) c^{2}

N = N_{o}e^{ - t }T_{1/2}=^{0.693}/^{}

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**
**

a)

b)

c)

To find time t_{B} we just use the
definition of velocity

t

t

Of course, we also have

Now we can apply the Lorentz Transformations to
find x_{A} and t_{A},

Now we can find the velocity in the A-frame and
determine if the Lorentz Transformations __work__. Is it, indeed,
__true__ that the velocity of light will be the same in both
frames? Let’s see, . . .

And, indeed, this ** is** the speed of
light!

This provides a more direct application of our velocity transformation equation,

with

v

f = c

f = c /

E = h c /

E = (4.14 x 10

E = (4.14 x 10

E = 2.26 eV

= (6.626 x 10

= 4.85 x 10

This wavelength is larger about a nuclear diameter
and is a reasonable fraction of typical atomic spacing.

Balmer Series for Hydrogen:

R = 1.097 x 10

For n = 3 we have

= (1.097 x 10

= 1.525 x 10

and this is
** red**
light

a)

b)

c)

E

E

E = hf = h(c/)

= hc/E

= (6.626 x 10 ^{- 34} Js)(3.0 x 10 ^{8} m/s) / 2.22
eV

We need to convert E = 2.22 eV into units of J (or the equivalent) to finish carrying out this calculation

E = 3.55 x 10

= (6.626 x 10

= 5.60 x 10

= 560 x 10

This is
** yellow**
light, near the middle of the range of visible light.

(C) 2002, Doug Davis; all rights reserved