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9 If the binding energy Per nucleon in $_3^7Li$ and $_2^4He$ nucler is 5.6 NeV and 7.06 MeV respectively, then in the reaction $P+_3Li -->2 (_2^4He)$ (P here retrent Proton) energy of Protpn must be
Energy of proton $(_1H^1)$ = = 4 (7.06)2 - 7(5.6) $= 2 \times 28.24 - 39.2$ =56.48-39.2 =17.28
9 f mo is the mass of an isotope$_3^{17}O$ , mp and mn are the masses of a Proton and neutron respectively, the binding energy of the isotope is
$B.E = \triangle mc^2 = (mo-8mp-9mn) C^2$
In gamma ray emission form a nucleus
In gamma ($\\gamma$) ray emission, the nucleus releases energy in the form of a gamma photon. This process does not change the proton number or the neutron number of the nucleus. Hence, the correct answer is that there is no change in the proton number and the neutron number during gamma ray emission.
The half life time of a radioactive elements of x is the same as the mean life of another radioactive element Y. Initially they have same number of atoms, then
$ ( T 1/2)_x = {0.693 \over \lambda x} (T)_Y = {1\over \lambda y} ({T_1 \over 2 })_x= T_y $ ${0.693 \over \lambda x} = {1/ \lambda y}$ $\therefore \lambda y = {dx \over 0.693 }$ $({T_1 \over 2 })_X= (T)_y $ $ \lambda y = 1.44 \lambda x $ $ {0.693 \over \lambda x} = { 1 \over \lambda y}$ $ According N = No\,e^{-\lambda t }$
The speed of daughter nuclei is
$ {1 \over 2} Mv^2 = ( \triangle m)C^2$ $ V^2 = {2(\triangle m) C^2 \over M}$ $V = C {\sqrt{2(\triangle m)} \over m}$
A and B are two radioactive substance whose half lives are 1 and 2 years respectively. Initially 10 g of A and 1 g of B is taken. The time after which they will have same quantity remaining is
$ {m_1 \over mb_1} = ( {1\over 2}) ^{1\over T1/2} = ({1 \over 2}) ^{t \over 1} = { 1 \over 2^t}$ $ m_1 = {10 \over 2^+} $ $ m_1 = m_2 $ $ m_2 = {1 \over 2^{t /2}}$ ${10 \over 2^t } = {1 \over 2^{t/2}}$ $ 10 = (2)^{t/2}$ $ log 10 = t/2 log 2 $ $1.0000 = {t \over 2} \times 0.3010 $ $ t =6.64 year \approx 6.6 year $
which of these is a fusion reaction
$_1^3H +_1^2H = _2^4 He + \,_0n^1$
The activity of a sample of a radio- active material is at time $t_1$ and $ A_2$ at time $ t_2$ (where $ t_2 > t _1 $) if T its mean life is then
$A_1 = A_0 e^{-t_1/t}$ $A=A_0^ {e -\lambda t } = A = A_0e^{-t/t}$ $A_0 = {A_1 \over e-t_1/T} \Rightarrow A_0 = {A_1 e^{t_1 /T}} $ $ Now , A_2 = A_0e^{-t_2/T}$ $ =A_1 e^{t_1/T} e^{-t_2/T}$ $ A_2 = A_1 e(t_1 - t_2)/ T$
In the following dis-integlation series $ _{92} U ^ {238} \rightarrow x \rightarrow _z y^A$ The value of Z and A respectively will be
$_{92}U^{238} \rightarrow x \rightarrow _zY^A$ $_{92}U^{238} \rightarrow _{90}Th^{234} \rightarrow _{91}Y^{234}$
gf $_{92}U^{238} undergoes succesively $8 \alpha$ decays and $6 \beta$ decays then resulting nucleus is
$_{92}U^{238} \rightarrow _{76}X^{206} \rightarrow _{82}Y^{206}$ $ \therefore _{82}Pb^{206}\; _{or}Pb^{206}$
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