Question

calculate Q and compare the following...NEED VERY URGENTLY WITHIN 1 HOUR THATS WHY I AM PROVIDING T00 MUCH POINTS...HAVE TO SUBMIT IN EXAM.....PLEASE SOLVE FULL WITH DETAILED EXPLAINIATION

The radionuclide C11 decays according to 6C11 rightarrow 5 B11 + 1 e 0 + v (T 1/2 = 20.3 min) The maximum energy of the emitted positron is 0 960 MeV. Give the mass values: m(6C11) = 11 011434 a.m.u., m (6 B 11) = 11 009305 a.m.u. Calculate Q and compare it with the maximum energy of the positron emitted.

Answer 1

Answer 2

Q = (m(6C11)-m(6B11))c^2
=(11.011434-11.009305)(931)
= 1.952099MeV

Answer 3

Answer 4

Change in mass = mass defect = (11.009305 + 0.00548) - (11.011434) = 0.003351 amu

Now each amu will corresponds to enerfy of 931.5 Mev

Hence enery Q = 931.5MeV * 0.003351 = 3.1214 MeV

It is approximately 3.5 times the maximum energy of the emitted positron

Answer 5

Answer 6

Since you need the answer to this question very urgently please provide more points at least another 3000 points!!

Answer 7

1 a.m.u = = 1.66053892e-27 kilograms

del m = 11.01144-11.009305 = 2.135*10^(-3) a.m.u

del m = 2.135*10^(-3)*1.66*10^(-27) kg = 3.544*10^(-30) kg

Q = delta m *c^2

Q= 3.544*10^(-30)*(3*10^8)^2 = 3.1897*10^-13 V

but 1 eV = 1.6*10^(-19)

Q = 3.1897*10^(-13)/1.6*10^(-19) = 1999125 eV = 1.999 MeV

Q obtained is greater then 0.96 MeV

Answer 8

The given nuclear reaction is:

$NS_4-11-08_Sravana_12_Physics_13_31_NRJ_$

Atomic mass of $NS_4-11-08_Sravana_12_Physics_13_31_NRJ_$ = 11.011434 u

Atomic mass of $NS_4-11-08_Sravana_12_Physics_13_31_NRJ_$

M aximum energy possessed by the emitted positron = 0.960 MeV

The c hange in the Q -value ( ? Q ) of the nuclear masses of the $NS_4-11-08_Sravana_12_Physics_13_31_NRJ_$ nucleus is given as:

$NS_4-11-08_Sravana_12_Physics_13_31_NRJ_$

Where,

m _e = Mass of an electron or positron = 0.000548 u

c = Speed of light

m

Answer 9

Q=(mass of reactants-mass of product)*c^2

where c is velocity of light.

so, change in mass= ((11.011434*1.66054e-27)-(11..009305*1.66054e-27)-9.11*10^-31)

so, Q=((11.011434*1.66054e-27)-(11.009305*1.66054e-27)-9.11*10^-31)*(3*10^8)^2

Q=1.4736Mev

so, it's energy is .5136Mev larger than that of positron.

Answer 10

Change in mass is Mass of carbon-(Mass of boron+Mass of electron+Mass of positron)=.001031 amu.

So Q= 0.001031*931.5 MeV=0.960375 MeV, which is same as of the maximum energy of positron.