1)
The
sketch of the decay curve for Problem (1) is shown above, and we note it
conforms to the half life (T½) = 6h, already computed.

2) The decay curve is shown for Problem 2 (accompanying sketch graph two above) and we see on inspection that since the counts per minute decreases from 360 to 180 in ONE half life (e.g. 1(T½)). The critical aspect to note is that the time is given in units of HALF LIVES not hours! Thus, since 1(T½)= 30 mins. = 1800s then (T½)= 1800 s. This is confirmed from the curve since the 180 counts/min decreases to 90 in 2 half-lives, and this decreases to 45 in 3 half lives and so on.

The activity A = ln 2/(T½) = 0.693/ 1800s

= 3.85 x 10

2) The decay curve is shown for Problem 2 (accompanying sketch graph two above) and we see on inspection that since the counts per minute decreases from 360 to 180 in ONE half life (e.g. 1(T½)). The critical aspect to note is that the time is given in units of HALF LIVES not hours! Thus, since 1(T½)= 30 mins. = 1800s then (T½)= 1800 s. This is confirmed from the curve since the 180 counts/min decreases to 90 in 2 half-lives, and this decreases to 45 in 3 half lives and so on.

The activity A = ln 2/(T½) = 0.693/ 1800s

= 3.85 x 10

^{-4}s
3)A = 5.6 x 10

^{-7}/s.(T½) = ln 2/ A = 0.693/ (5.6 x 10

^{-7}/s)

(T½) = 1.24 x 10

^{6}s = 14.3 days

4)

*Solution:*

Given: dN/dt = - lN

So
that:

*l***= 1/N****|**dN/dt**|****= 10**^{-15}(6.00 x 10^{11}/s)**l**= 6.00 x 10

^{-4 }s

^{-1}

A
= = - ( 6.00 x 10

6.00 x 10

*-**l**N*^{-4 }s

^{-1})( 10

^{15})= - 6.00 x 10

^{11}/s i.e.

^{11 }

*decays per second*

c) Half life:

0.693/ (6.00 x

**T**= ln 2/l = 0.693/ l =_{½}0.693/ (6.00 x

^{ }10^{-4 }s^{-1})
= 1160 s
(or 19.3 minutes)

5)

**(a): we note that for a multi-electron atom we have for the Bohr –atom energy:***Solution*
E

_{n}= - 13.6 (Z_{eff})^{2}/**n**^{2 }**where:**^{ }**Z**_{eff}_{ }= (Z – 10)
Note
that

**Z**_{eff}_{ }is the atomic number (Z =11) minus the number of electrons between the nucleus and the electron being considered. Since 10 electrons separate via ionization the last remaining sodium electron (e.g. the electron that occupies the 3s sub-level of the n= 3 principal level (electron arrangement for sodium is: (1s^{2}, 2s^{2}, 2p^{6}, 3s^{1})).
Then
(given the 1

^{st}line corresponds to n = 3):
E

_{n}(eV) = - 13.6 (Z- 10)^{2}/**(3)**^{2}

^{ }**E**

^{ }_{n}= -13.6 eV (1)/9 = 13.6 eV/9 = 1.51 eV

Now,
the energy is related to the wavelength (

*l***in nm by:***)*

*E*

*l*

*= 1.99 x 10*^{-16}J nm
To
get the energy in Joules, we use the fact that:

1
eV = 1.6 x 10

^{-19}J so 1.51 eV = 1.51 eV(1.6 x 10^{-19}J /eV)
E
= 2.41 x 10

^{-19 }J Then:

*l***1.99 x 10**

*=*^{-16}J nm

**2.41 x 10**

*/*^{-19 }J = 825 nm

(b)
The
voltage of 10

^{4}V leads to an energy:
E
= qV = (1. 6 x 10

^{-19}C) (10^{4}V) = 1. 6 x 10^{-15}J
6)

**Solution****:**The activity of a radio-nuclide is given as:

*A = A*_{o }exp (-

*l*

*t)*

Where

**is the decay rate at time t = 0, and***A*_{o}**refers to the decay rate at some time t thereafter.***A*
a)
The decay constant

*l***= 0.693/****T**=_{1/2}
0.693/ 883612800s

Or:

*l***=****7.84 x 10**^{-10}^{ }
b)
The activity

**after 1 hour (3600 s) is given by:***A*
A = A

_{o }exp (-lt) = 1. 1 x 10^{10}decays/sec (exp (-lt) )
Where (lt)
= (

**7.84 x 10**^{- 10}^{ }) (3600 s) =**2.82 x 10**^{- 6}
Then:

A

_{o }exp (-lt) = 1. 1 x 10^{10}/s [exp (-**2.82 x 10**)] =^{- 6}
1.
1 x 10

^{10}/s [1.00000] = 1. 1 x 10^{10}decays/sec
After
two hours (t = 7200s ):

(lt) = (

**7.84 x 10**^{- 10}^{ }) (7200 s) =**5.64 x 10**^{- 6}_{o }exp (-lt) = 1. 1 x 10

^{10}/s [exp (-5

**.64 x 10**)] =

^{- 6}
1. 1 x 10

^{10}/s [1.0000] = 1. 1 x 10^{10}decays/secAfter 49 years, T = 1. 54 x 10

^{9}sec

lt = (

**7.84 x 10**

^{- 10}^{ }) (1. 54 x 10

^{9}s) =

**1.20**

A

_{o }exp (-lt) = 1. 1 x 10

^{10}/s [exp (- 1.20)]

=1. 1 x 10

^{10}/s (0.301) = 3.31 x 10^{9}decays/sec
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