Find the energy liberated in the reaction 223Ra - 209Pb - 14Cthe atomic masses needed are as follows 223Ra 209Pb 14C 22-018u 208-981 u 14-003 u

nbsp; ^223R_a = 223.018 u, ^209Pb= 208.981 u; ^14C = 14.003 u ^223R_a → ^209Pb + ^14C m = mass ^223R_a mass ( ^209Pb +^14C ) → = 223.018 ...

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Standard XII

Chemistry

Stoichiometry

Find the ener...

Question

Find the energy liberated in the reaction
$^{223} R a \to^{209} P b +^{14} C$
the atomic masses needed are as follows
$^{223} R a^{209} P b^{14} C$
$22.018 u 208.981 u 14.003 u$

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Solution

$^{223} R_{a} = 223.018 u,^{209} P b = 208.981 u;^{14} C = 14.003 u$
$^{223} R_{a} \to^{209} P b +^{14} C$
$△ m = m a s s^{223} R_{a} - m a s s (^{209} P b +^{14} C)$
$\to= 223.018 - (208.981 + 14.003) = 0.034$
$e n e r g y = △ M \times u = 0.034 \times 931 = 31.65 M e V$

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Similar questions

Find the energy liberated in the reaction: $^{223} R a \to^{209} P b +^{14} C$
The atomic masses needed are as follows:
$^{223} R a 223.018 u$ $^{209} P b 208.981 u$ $^{14} C 14.003 u$

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Find the energy liberated in the reaction223Ra→209Pb+14Cthe atomic masses needed are as follows 223Ra209Pb14C22.018u208.981u14.003u

223Ra=223.018u,209Pb=208.981u;14C=14.003u223Ra→209Pb+14C△m=mass223Ra−mass(209Pb+14C)→=223.018−(208.981+14.003)=0.034energy=△M×u=0.034×931=31.65MeV

Find the energy liberated in the reaction
$^{223} R a \to^{209} P b +^{14} C$
the atomic masses needed are as follows
$^{223} R a^{209} P b^{14} C$
$22.018 u 208.981 u 14.003 u$

$^{223} R_{a} = 223.018 u,^{209} P b = 208.981 u;^{14} C = 14.003 u$
$^{223} R_{a} \to^{209} P b +^{14} C$
$△ m = m a s s^{223} R_{a} - m a s s (^{209} P b +^{14} C)$
$\to= 223.018 - (208.981 + 14.003) = 0.034$
$e n e r g y = △ M \times u = 0.034 \times 931 = 31.65 M e V$