ionistaion energy of H atom and energy released

Jun 2014
306
0
for part a , the question want ionistaion energy ... why shouldnt it be H2 atom dissciate to become 2H+ ion? but the solution gives H+ + e- = H instead?

for part c , shouldnt the loss of mass from 10.0kg of LiD= 0.002988x(10^6)x10= 29.88kg?

then use E= mc^2 to proceed?
 

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topsquark

Forum Staff
Apr 2008
3,055
651
On the dance floor, baby!
for part a , the question want ionistaion energy ... why shouldnt it be H2 atom dissciate to become 2H+ ion? but the solution gives H+ + e- = H instead?

for part c , shouldnt the loss of mass from 10.0kg of LiD= 0.002988x(10^6)x10= 29.88kg?

then use E= mc^2 to proceed?
As we cannot see parts a) and c), much less the original question there is no possible way to give you an answer.

As a guess on c), I'm assuming that we are using the loss of mass of 10.0 kg of LiD from b) part iii), in which case where did the 10^6 come from? And anyway, look at the numbers: You are saying that the loss of mass of 10.0 kg of LiD is 29.88 kg, 19.88 kg more than you started with!

-Dan
 
Jun 2014
306
0
As we cannot see parts a) and c), much less the original question there is no possible way to give you an answer.

As a guess on c), I'm assuming that we are using the loss of mass of 10.0 kg of LiD from b) part iii), in which case where did the 10^6 come from? And anyway, look at the numbers: You are saying that the loss of mass of 10.0 kg of LiD is 29.88 kg, 19.88 kg more than you started with!

-Dan
 

topsquark

Forum Staff
Apr 2008
3,055
651
On the dance floor, baby!
for part a , the question want ionistaion energy ... why shouldnt it be H2 atom dissciate to become 2H+ ion? but the solution gives H+ + e- = H instead?
You can do that but you have another energy release. H_2 becoming 2H implies a loss of binding energy from the H_2 nucleus. You would have to calculate that, then you would have the chained reaction:
H_2 --> 2H + binding energy --> 2H+ + binding energy + 2(13.6 eV)

However since all you want is the mass loss between H and H+ your extra efforts are wasted as all we care about is the last two parts in which the binding energy plays no part.

-Dan