Friday, October 24, 2014

Millennials Need to Stop Whining About the Boomers!

Let's face it, the Baby Boomers have been a target of easy opportunity for all manner of nattering nabobs of negativity, who insist we Boomers are about to usher in the fall of civilization as we know it. From Arch-Boomer nemesis, Robert Samuelson who blames “entitlements” for an approaching fiscal train wreck, to a spate of authors – including P.J. O'Rourke, Joe Klein and  David Willets (in his book, ‘How the Baby Boomers Took Their Children’s Future and Why They Should Give It Back’)

Now enter a whiny Millennial, Alexander S. Balkin, who in a recent Salon piece (‘Boomers Ruined America) cried for over 1,000 words on how terrible, amoral and  greedy the Baby Boomers are with choice shots such as:

The perfect example of this was the 2008 collapse of the toxic housing debt market. In government, baby boomers ballooned the defense budget beyond the point of reason. They then raided government programs to pay for their mistakes. Regarding the environment, baby boomers left the United States reliant on coal (cough, cough) while eroding the advanced nuclear energy infrastructure built by their parents. We can thank baby boomer leadership for a nation that has no sound policy on foreign affairs, the environment, energy, social welfare, human rights, terrorism, technology development, education, debt, etc. The point being, baby boomer leadership has provided America with a government that is the most partisan and self-serving the union has ever seen, and remains entirely reactive to the world around it.
All of which is over-generalized mush – can you say “fallacy of composition” i.e. when one claims that what is true of a part is true of a whole?  In this case, the “Boomers” he is referring to make up about 5 percent of the total generation. These are the entitled, 'silver spoon in the mouth'  twerps like Gee Dumbya Bush – who got his Harvard education via a legacy entitlement, and who – once elected- set the nation on a course of perpetual fear (following 9/11) , perpetual war (with defense budgets to match)  and massive oil  & coal dependency (well, BOTH Bush and Cheney were oilmen, after all, duh!)

But this twerp Balkin makes no distinction between that enclave of miscreants and the vast mass of Boomers – like me – who gave blood, sweat and tears either fighting in Vietnam for a war those like Bush instigated (in that case, LBJ – fawning fool for the Texas oilmen and military industrial complex) or serving in Peace Corps (as I did for 4 years) or some other volunteer mode. Yet this guy blames us and essentially dumps us all into the same mix as the Bushite- Neocon- Maul Street shyster  subset of Boomers!  This lack of rational discrimination alone calls into question his credibility in writing the piece.

As for the 2008 financial meltdown, don’t even get me started. But if this guy Balkin had a clue he could have researched and excavated the actual background – which can be boiled down to those of the elite Bushie-mold ilk creating credit default swaps and mixing them up with valid bonds and labeling the whole AAA. Those who want more information (and that especially includes Mr. Balkin)  can go here:


Also, the historical causes of multi-nation financial meltdown are a lot less obvious than picking on one highly visible demographic.  If you want the supreme distillation it can basically be summed up in two parts: 1) trashing the Bretton Woods Global Finance Agreement (1973), and 2) Launching numerous wars of choice which have not only cost blood but enormous treasure.

In the case of Bretton Woods, while it wasn’t perfect it did provide a more equitable global basis for competition and trade than what’s in place now – based on such travesties as G.A.T.T., NAFTA and World Trade Organization fiats (which sent millions of American jobs overseas). Worse, with the collapse of Bretton Woods, global currencies went from a fixed exchange rate system to a floating system. Only years after did many of the creators of the latter realize what an error they had made, and how much they facilitated currency speculation –especially in the carry trade. Since the Asian crisis in 1997, currency raiders have splattered the reserves of a number of nations as governments have watched helplessly. Only most recently, this piracy has reached new heights via the used of “naked credit default swaps” for speculators to bet on national collapse (as in Greece) with NO money on the table of their own.

If one now factors in how things have changed since 1973, one begins to see why the “younger generation” has been placed in economic peril – and it isn’t from the Boomers! For one thing, floating exchange rates and currency raids have allowed the dollar’s strength to deteriorate. This is called “currency debasement” and it’s been used over and over in recent decades for not very salutary purposes. For example, the Bush government used it to make the cost of domestic goods cheaper (via a cheaper dollar) to try to reduce the U.S. trade deficit.

Then again, the global market tyranny that has replaced Bretton Woods, has enabled millions and millions of high paying, relatively secure jobs (especially in manufacturing, but more recently in computer software specialties) to be shipped offshore to places like Bangalore and Beijing. Minus those higher benefits, higher paying jobs, Americans have barely been able to make ends meet – and this partly explains their stupendous reliance on credit card – to make up for their lost income. Young people, caught in this vortex, will obviously not see the same economic security as their parents who held steady good paying jobs with top benefits, to the extent that one could even afford to stay home. The now (nearly extinct) single income family!

Now, to this, add in all the wars  (Bushie instigated- without higher taxes to pay for them) which have bled down the national treasury (as I noted by an estimated $6 trillion – not even counting whacked out “defense” spending such as the $2.1 trillion under Reagan) and you have the basis for an economic Armageddon. Because make no mistake, SOMEONE had to pay or will have to pay, for those wars. Who is holding most of the bonds right now? Chinese and Japanese bankers – to the tune of nearly $2 trillion. Who will get the bill in the future? Our kids and grand kids, because their cowardly government didn’t have the brains or insight not to try to act like the world’s cop – and especially (in the Bush years) launching a full scale war against robed lunatics when a localized police or intel action could have been effective.

They preferred to bankrupt us, as Osama bin Laden always wanted – playing into his hands. Nor did they have the inner fortitude, given they got their war vote (by reconciliation no less) , to at least mandate the taxes to pay for it.  They preferred instead to use future taxes (and higher interest rates) to pay for it, but on the backs of future generations.

Yet now the smarmy asses in congress that voted for these war travesties and their humongous costs, want to say we haven’t the money for decent health care reform.

No, the Boomers are not the ghouls they’re made out to be by the likes of David Willet  or Alexander S. Balkin.  One of whom (Willet)  blames us for having too few children to increase our wealth relative to theirs and the other blames us for everything from the credit crisis to exploding defense budgets and overt dependence on fossil fuels.  Neither displays any grasp of reality and therein lies the problem. Because minus test in reality and hard evidence you don't know what you're talking about. 

People need to wake up, stop their blame game, and examine who the real culprits are in this forlorn multinational debt and war miasma. One thing for sure: Boomers do not need to “give back” anything to the next generation. What we do have to offer is our insight, and best advice. It’s up to the young ‘uns to take it or leave it, but if they leave it – don’t blame us for their predicament and lack of political smarts.

If I had one piece of advice for the likes of Balkin and his compatriots, it would be to put down the Ipods, smart phones and Twitter access and head to the library to do some serious reading: on history, politics and economics especially. To see first hand how you got to where you are without having to rely on simplistic scapegoats....like a whole generation. Knowledge is the key to your liberation, not picking on a large group that in many ways has been victimized as much or worse than you and for a longer time.

And btw, despite Balkin's claims we were somehow all privileged and got it all handed to us, I will have him know that most of us weren't favored sons like George W. Bush. We either had to get scholarships to attend university or student loans. And after graduating we educated ourselves on the basics of the political system and economics - our parents didn't have the time to dote because they were too busy working.

Before Mr. Balkin or others of his generation go after the Boomers again, they may wish to pick up a copy of my book, The Elements of the Corporatocracy – to see how Millennials really ended up where they are:


Introduction to Nuclear Physics (3): Nuclear Fusion and Deuteron Potential

Continued from previous installment:

4. Nuclear Fusion Reactions:

In general a nuclear fusion reaction is one in which two light nuclei combine (fuse) to form a heavier nucleus with positive energy given off (the Q of the reaction).  Nuclear fusion is demonstrated in its most compelling form in the case of stellar energy. Exhaustive investigations in this regard, eventually led to the realization that fusion was the only practical energy by which stars could be sustained over long periods of time, such as billions of years.

In the Sun, for example, two distinct nuclear fusion processes occur: 1) the proton-proton cycle, and 2) the carbon-nitrogen cycle.

 In the first of these (the easier one because it has fewer reactions):
                                                                                                                       
1H + 1H + e- ®  2 H   + n + 1.44 MeV

2 D   + 1H ®  3 He + g + 5.49 MeV

3 He + 3 He ®  4 He + 1H + 1H  + 12.85 MeV

The top line shows two protons fusing to yield deuterium (heavy hydrogen) with a positron and neutrino (n) emitted, along with 1.44 MeV of energy. Empirical evidence of this reaction is obtained from gallium detectors, of the neutrinos given off, which are within 1-2% of what theoretical models predict.[1] In the second fusion reaction, the deuterium combines with a proton to give the isotope helium 3, along with a gamma ray (g) and 5.49 MeV energy. In the final fusion, two helium-3 nuclei combine to yield one helium-4 nucleus, along with two protons, and 12. 85 MeV energy. Note that the two ending product protons commence the cycle anew, so that the generation of nuclear energy is ongoing.  

The ending quantities on the right sides of each part of the cycle denote the Q of the reaction for that part.  Let us check the Q for the first and simplest part. We know the hydrogen mass = 1.007825 u and for deuterium we have (from atomic tables): : 2 D  =  = 2.01410 u. Then:

Q = [ 2(1.007825 u) – 2.01410 u] c2

Q = [ 2.01565  u – 2.015941u] c2

Q = [2.01565 – 2.01410] 931.5 MeV/u

Q = [0.00155] 931.5 MeV/u = 1.44 MeV

The effect of ongoing fusion reactions such as this,  means that the central core of the Sun becomes heavier and heavier, as more and more helium is produced. This despite the fact that the Sun as a whole is losing an amount of mass of roughly 4 x 106 metric tones per second

Insight Problem:

If the atomic mass for helium 3 (3 He) is equal to 3.01603 u, then verify the other Q-values for the last two parts of the proton-proton cycle. A  simplified, compressed “net reaction”:

 1H + 1H +1H + 1H ®  4 He + Energy

Is sometimes used to evaluate the total energy released in the proton-proton cycle. Compute this energy and compare to the value obtained for the total energy released in the earlier example. Can you account for the difference?

Nuclear Fusion Reactions in the Aging Sun:

     At some stage, when nearly the entire solar core is helium a new helium fusion phase will be ushered in (at higher temperature), such that the following reaction series, known as the ‘triple alpha’ process, kicks in:

4He + 4He ®  8Be + g  (- 95 keV)

8Be + 4He  ®  12C + g  + 7.4 MeV

     Here, the two alpha particles (helium nuclei) first fuse to give unstable beryllium and a gamma ray (g), with 95 keV energy absorbed. Then the beryllium fuses with a helium-4 to give carbon–12 plus a gamma ray and 7.4 MeV energy given off.

     In this way a new cycle commences, leading to a heavier molecular weight core. Each successive burning phase, however, is less efficient than its predecessor, as can be seen by comparing the energy given off in the triple alpha process to the energy given off in the proton-proton cycle. The key thing to bear in mind in terms of a stable phase (i.e. ‘Main sequence’) star like the Sun is that it is in pressure-gravity balance. The outer gas pressure balances the weight of its overlying layers. Any condition likely to disrupt this balance is therefore of paramount interest.

The stable lifetime of the Sun depends on how long before it consumes ninety percent of the hydrogen in its core. Theoretical investigations using data from nuclear reaction rates and cross sections suggest the Sun’s Main Sequence lifetime at 8-10 billion years. Since it already has spent 4.5 billion of those years, there are anywhere from 3.5 to 5.5 billion years remaining. 

Once the triple-alpha process gets underway and the energy balance declines, the Sun will have to compensate for the lost energy to sustain any kind of balance. Thus, the Sun’s core must contract and convert gravitational potential energy into thermal energy.  Meanwhile, ignition of hydrogen burning in the Sun’s outer layers will create radiation pressure that forces the outer layers to expands. The Sun will then become a “Red Giant” and its new larger surface will be expected to engulf all the planets up to and including Mars.

Example Problem:

If the atomic mass of beryllium 8 (8Be) = 8.00531 u, verify that the first part of the triple-alpha fusion process is endothermic and has the value given.

Solution:

We have:

Q = [ 2(4.00260 u) –  8.00531 u] c2

Q = [ - 0.00011] 931.5 MeV = 0.102465 MeV = - 102.4 keV

Of course, not taken into account here is the gamma ray (g) which also comes off. Hence we will have:

(-102.4 keV) + (E (g)) = -95.7 keV

So that:

E (g) =   hc/ l = 6.7 keV

Is the missing energy of the gamma ray photon, with the difference factored in yielding 95.7 keV.


Aside: The Problem of the Coulomb Barrier in Solar Fusion

 The problem of the Coulomb repulsive barrier to solar nuclear fusion was first highlighted and explored by Prof. Martin Schwarschild in his excellent monograph 'The Structure and Evolution' of the Stars’  (Dover, 1958).

     Schwarzschild once calculated that the probability of any one proton fusion in the Sun’s core would ordinarily be  about once every 14 billion (14 x 109) years. Since the universe itself is only 13.8 billion years old this means it could never occur unless another factor was present to enable it.

     The reason for this has to do with the Coulomb (electrostatic) repulsion between the potentially fusing H-nuclei. Thus, each proton, having (+) charge tends to repel any other proton within a discrete sphere or distance around it. (Recall from your basic physics, like charges repel, unlike attract - and that's what essentially obtains here)

     In order for thermonuclear fusion to be realized, the Coulomb barrier must be overcome. Fortunately quantum mechanics allows for a certain non-vanishing probability that a particle (say proton) of kinetic energy K, can overcome a barrier of energy V ("barrier potential"), via the process of "quantum tunneling".

Note that tunneling is a general feature of low mass systems, such as single proton (H) states.

Consider a deBroglie (matter) wave arising from a single proton (p+) of form:

U(x) ~ sin(kx)

Where x is the particle's linear displacement (e.g. in 1-D) and k, the wave number vector(k= 2π/
l), where l  denotes the wavelength.

    Though the associated kinetic energy K < V (the barrier "height") the wavefunction is *non-zero* within the barrier, e.g.

U(xb)~ exp(- cx)

So, visualizing this behavior as shown below: 

Fig. 4: Tunneling through Coulomb barrier potential  to allow nuclear fusion


with the "barrier" at height V, so we can visualize the particle of lesser energy K, moving from the left side of the E-axis "tunneling" through to the right side where it may have wave function, U(x) ~sin (kx + φ), where φ denotes a phase angle.

    Note that if the barrier is not too much higher than the incident particle energy, and if the mass is small, then tunneling is significant.

    It's important here to point out that the penetration of the barrier is a direct result of the wave nature of matter. In effect, this wave nature - which is uniquely quantum mechanical in origin- allows a higher energy barrier to be penetrated by a lower energy particle, something totally without parallel in classical, Newtonian physics!

    Even given tunneling, an "offset" is required to reduce the low penetration probability , since clearly the Sun and other stars are shining by fusion.

     This 'offset' arrives via enormously high density of protons, e.g. in the core, which: i) increases the probability enormously, since so many more protons are in extremely close proximity, and (ii)enhances temperatures to the point they can be sustained, and continue - thereby building up other fusion reactions to finish the initial one.

   The idea here being that a particle of relatively low incident energy (of kinetic energy K, say) can actually penetrate a higher potential energy barrier, say of energy V(x) > K. Note that the penetration of the barrier is a direct result of the wave nature of matter! (The matter wave form changes in the process of transmission through the barrier, say from an exp(-ikx) function to a sin (kx + f) where f denotes phase angle). In effect, this wave nature - which is uniquely quantum mechanical in origin- allows a higher energy barrier to be penetrated by a lower energy particle, something totally without parallel in classical, Newtonian physics! Note that if the barrier is not too much higher than the incident energy, and if the mass is small, then tunneling is significant.   It was insights such as this that paved the way to apprehending how much subtler nature was than hitherto realized, and how many more technological advances could be achieved when the wave nature of matter was factored into designs. 


5. The Quantum Treatment of the Deuteron.

The deuteron is perhaps the most basic nuclear system to confront. As we know the deuteron consists of one proton, one neutron and the electron – with the first two comprising the nucleons.

To proceed, we write the usual Schrodinger equation for the hydrogen atom but let F and Q = const. so that their derivatives are zero. Then, substitute in the reduced mass:

m’=   m n m p/ (m n + m p)

So we obtain the new form of the Schrodinger equation:

1/r2 d/dr (r2 dR/dr)  + 2m’/ ħ[E – V] R = 0

This can be further simplified by letting:

U(r)  = rR(r)    which results in the equation:

d2U/dr2  +  2m’/ ħ[E – V] U = 0


In this equation we find that V, the potential, has two values:

V = - Vo and  V = 0 outside the well. The diagram below shows how we are treating the deuteron in terms of the function V(r).



Fig. 1: The potential well associated with deuteron

There are then two solutions we can designate:

i) u I for r   <   r o and 

ii) u II for r   >   r o

Inside the potential well:

d2 u I /dr2  +  2m’/ ħ[E + Vo] u I = 0

and we let:   a2  = 2m’/ ħ[E + Vo]

so that: 

d2 u I /dr2  +  au I = 0


For which we can show:

u   = A cos (ar)  + B sin(ar)

Since R = u/ r  the cosine solution must be discarded, lest we get an unwanted infinity. This leaves:

u   =   B sin(ar)

Outside the potential well  V = 0 so that:

d2 u II  /dr2  +  2m’/ ħ[E]  u II = 0

Let:

b2  = 2m’/ ħ[- E ]


Since the total energy of the neutron is negative, i.e. being bound to the proton.  Then:

d2 u II  /dr2  -  b2 u II = 0

Which can be shown to have the solution:

u II   = C exp (-br) + D (exp(br)

For consistency we demand u ® 0 as r ®¥, so D = 0 and:

u II   = C exp (-br)

For continuity at r  =   r o  and  u I  =  u II  :

B sin(ar)  =  C exp (-br)

Thence:

du I /dr = d u II  /dr Þ  aB cos (a r o) = -bC e-b r o

Now, divide the solution on the left side by the one on the right side, e.g.:

A B cos (a r o) /   -b C e-b r o

so:

tan (a r o)  = -  a/b

In effect, the deuteron problem cannot be solve analytically only graphically.


Quantum Numbers for Deuteron:

Since there are two particles each with intrinsic spin ½   in the deuteron, the total intrinsic spin angular momentum can only have the values: S = s1 + s2 = 0 or 1.   The orbital angular momentum quantum number, L (describing the motion in space of the proton and neutron relative to each other) can assume the values L= 0, 1, 2 (i.e. S, P, and D states).

The total angular momentum of the deuteron has been measured and the total angular momentum quantum number J has been found to be 1. This must be the vector sum of the orbital angular momentum (L) and the total spin momentum (S).


Problems:

1) Calculate the wavelength of the gamma ray photon (in nm) which would be needed to balance the endothermic part of the triple –alpha fusion equation. (Recall here that 1 eV = 1.6 x 10 -19 J)

2) Verify the second part of the triple-alpha fusion reaction, especially the Q-value. Account for any differences in energy released by reference to the gamma ray photon coming off and specifically, give the wavelength of this photon required to validate the Q.

3) The luminosity or power of the Sun is measured to be L = 3.9 x 1026 watts.  Use this to estimate the mass (in kilograms) of the Sun that is converted into energy every second. State any assumptions made and reasoning.


4) In a diffusion cloud chamber experiment, it is found that alpha particles issuing from decay of U238 ionize the gas inside the chamber such that 5 x 10 3 ion pairs are produced per millimeter and on average each alpha particle traverses 25 mm. Estimate the energy associated with each detected vapor trail in the chamber  if each ion pair generates 5.2 x 10 18 J.

5) When 118 Sn 50 is bombarded with a proton the main fission fragments are: 

24 Na 11   and  94 Zr 40  

The excitation energy necessary for passage over the potential barrier is:

e  >   3 ke2 Z2/ 5

Where the right hand side denotes the height of the Coulomb barrier.

a) What must this value be?
(Take   ke2  = 1.44 MeV/ fm)

b) Find the energy difference between the reactants and the products. (Take c2  = 931.5 MeV/ u)


6)  a) Show that u   = A cos (ar)  + B sin(ar)

In  a solution of the reduced Schrodinger equation for the deuteron:

   d2 u I /dr2  +  au I = 0.

Show why the cosine solution needs to be discarded.

 
b) Show that u II   = C exp (-br)

Is a solution of the other reduced Schrodinger equation for the deuteron:

   d2 u II /dr2  +  bu II = 0? 

Explain.



[1] See, e.g. Physics Today: Reports, April, 1995, p. 19.

Thursday, October 23, 2014

Solutions to Problems - Intro. To Nuclear Physics (2)

1.Write out the full nuclear reaction for:

13 Al 27 (a, n) 15 P 30

Thence, or otherwise, find the Q of the reaction.

Solution: We write out the reaction from its condensed form given:

13 Al 27  + 2 He 4      ®  15 P 30 0 n 1    


Then: The Q of the reaction is defined according to:

Q = [Ma + MX  - MY – Mb]c2

Where:

0 n 1 = Mb     = 1.008665u

a = MX  =  2 He 4  =  4.002603u

15 P 30    = MY  =   29.9783138u  (from Wikipedia table, Googled)

13 Al 27  = M =  26.9815386u (from Wikidpedia table – Googled)

Q =   [26.9815386u  +   4.002603u  - 29.978313 u  - 1.008665u] 931 MeV/u


=   [30.984141 u  -   30.986978]u (931 MeV/u)= - 2.64 MeV

(Negative sign denotes an endothermic reaction)

2. Identify the missing ‘X’ in each of the following:

a) 84 Po 215   ®  X + a

b)   N 14 (a, X)  O 17

c) 48 Cd 109    +   X ®   47  Ag 109     +  u


Solutions:

a)     This is an example of alpha decay defined by:

Z X A ®  Z-2 X A-4 + 2 He 4

Then:  A – 4 = 215 – 4 = 211

And:  Z – 2 =  84 – 2 = 82

So: X =   Pb (lead) e.g.  82 Pb 211  

b) This is a condensed form for the reaction which would be written out:

7 N 14   + a ®  X + 8 O 17  

Or:  7 N 14    + 2 He 4     ® Z1 +Z2  X A1 + A2  + 8 O 17  

Now: A1 + A2 =  14 + 4 = 18 and Z1 + Z2 = 7 + 2 = 9

So: Z1 +Z2  X A1 + A2  =   9  X18  which is fluorine,  or X = F (e.g. 9  F 18 )


b)     By inspection we see the mass number Z remains unchanged but the atomic number decreases by one. This must be an example of beta decay, defined:


Z X A + -1 e0 ®  Z-1 X A   + u

Hence, X = -1 e0

So that: 48 Cd 109    +-1 e0   ®   47  Ag 109     +  u


3. Consider a process of neutron removal whereby:

20 Ca 43    ®   20 Ca 42     + 0 n 1    

We wish to find the “separation energyS n associated with the neutron.  Estimate the value of this energy if the atomic mass of Ca 43     = 42.98780u, and  Ca 42  = 41.958625u, and take the neutron mass = 1.008665u.

S n  is just another variant of the Q for a reaction where:

S n  =   [(total rest mass before decay) –
                                (total rest mass after decay )] c2

S n  =  

   [(42.98780 u) -  (41.958625u + 1.008665u)] 931 Mev/ u  = 

(42.98780 u – 42.96729u ) 931 Mev/ u  = 

 0.0205 u (931 Mev/ u ) = 19.09 MeV

4. For each of the following reactions, write out the full nuclear equation and find the Q of the reaction:

a) H 2 (d, p)  H   (Note: d is for deuterium or 1 H 2 )

b) Li 7 (p, n) Be 7


Solution:

a)  1H 2  +   1H 2  ®  1H +   1H 3  

Q = [Ma + MX  - MY – Mb]c2

1 H 2 = Ma     = 2.01410 u

1 H 2  = MX  =  2.01410 u

1H 1 = M = 1.007825 u

   1 H 2  = MY  =   3.01604 u (from Wikipedia)

Therefore:

Q = [2(2.01410 u)  - 1.007825 u - 3.01604 u] 931 Mev/u

=  [(4.0282) – 4.0238)] 931 Mev/u  =   0.0044 (931 Mev/u  ) = 4.09 MeV


5. Refer to the example of the fission of U 235 illustrated at the outset of Part 1. This fission reaction may be written:

92U 235   +  0 n 1     ®  42 Mo 95     +    57 La 139 + 2 (0 n 1)

Find the Q-value of this reaction, given these atomic masses:

   0 n 1     = 1.009u

92U 235   = 235.123u

42 Mo 95   =  94.945u

57 La 139   =  138.955u

Q = [Ma + MX  - MY – Mb]c2

Where:

92U 235    = Ma   = 235.123u

0 n 1       = MX  = 1.009u

42 Mo 95  +  57 La 139  = M =94.945u + 138.955u =  233.900 u

 2 (0 n 1)  = MY  =   2(1.009u) =  2. 018 u

Then:

Q = [(235.123u + 1.009u) – (233.900 u +2. 018 u)] c2

Q =   [(236.132)  -  (235.918)] 931MeV/u  =  0.214(931MeV/u)


Q =  199.2 MeV