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A gamma ray Photon creates an electron- Position Pair. If the rest mass energy of an electron is 0.5 MeV. and the total kinetic energy 0.7 MeV, then the energy of the gamma ray Photon must be
$Ionigation energy of atom E = {13.6 Z^2 \over n^2} ev$ $For helium z = 2= { 13.6 \times (2)^2 \over (1)^1 } = 54.4 ev$
Large angle scattering of $\alpha$ - particle could not be explained by
$mass \, defect \, \triangle m = ( 2mp +2mn) -MHe$ $ \triangle m = ( 2 \times 1.0087 -2 \times 1.0073) - 4.0015 $ = 4.032-4.0015 $ \triangle m = 0.030 amu$ $ E = \triangle m \times 931.48 mev$ $=0.0305 \times 931.48 = 28.4 MeV$
The energy of an eleforn in $n^{th}$ orbit of hydrogen is ${ -13.6 \over n^2 } eV$ energy required to exite the electron form the first orbit $4^{th}$ orbit is
Thomson model
The activity of a radioactive sample is measured as no counts Per minute at t=o and No counts Per minute t=5 min The time (in min) at which activity reduces to half its value is
$\triangle E= E_4 - E_1 = {-13.6 \over 4} - ({-13.6 \over 1})= 0.85 +13.6 = 12.75 ev$
A heavy nucleus at lest breaks into two fragments which fly off with velocities in the ratio 8:1 The ratio of radil of the fragments is
In mean life (T) Rest mass = $No/ e$ $ T = { 1 \over \lambda } \Rightarrow \lambda = {1 \over 5 } min ^{-1}$ $ N = N0e^{- \lambda t}$ $ N0/2 = N0e^{-1/5 t}$ $m2 = { 1\over 5}t $ $ {1 \over 2 } = e^{-1/5} t $ $ t = 5ln2$ $ 2^{-1} = e^{-1/5} t $ $ t = 5 log e^2$ $ 2^1 = e {1 \over 5 } t $
The Probability of survival of a radioactive nucleus for one mean life time is
In the nuclear decay below $ _ZX^A \rightarrow _{Z+1}Y^A \rightarrow _{Z-1}B^{A-4} \rightarrow _{Z-1}B^{A-4}$
$ _ZX^A \rightarrow _{Z+1}Y^A \rightarrow _{Z-1}B^{A-4} \rightarrow _{Z-1}B^{A-4}$
The wave lenght of the first line of Lyman series for hydrogen atom is equal to that of hydrogen atom is equal to that of second line of Balmar series for a hydrogen like ion. The atomic number Z of hydrogen like ion is
$ {1 \over \lambda } = R2^2[{1\over n_2^2} -{1\over n_1^2}] \leftarrow for \, any \,atom $
$ { 1 \over \lambda_1 (L) } = R(1)^2[{1\over 1^2} - {1 \over 2^2}]=RZ^2[{1\over 4} - { 1 \over 16 } ] {3RZ^2 \over 4}$
$ { 1 \over \lambda_1 (L) } = {3R \over 4} \Rightarrow \lambda_1(L) = { 4 \over 3R} $
$ { 1 \over \lambda_2 (B) } = R2^2[{1\over 2^2} - {1 \over 4^2}]=RZ^2[{1\over 4} - { 1 \over 16 } ] = {3RZ^2 \over 16}$
$ \Rightarrow \lambda _2 (B) = { 16 \over 3RZ^2} $ $ { \lambda_1 (L) \over \lambda _2 (B) }$ $ 1 = { 2^2 \over 4} \Rightarrow z^2= 4 Z=2$
A nucleus $_n X ^m$ emist one -$ \alpha $ Particle and two $\beta$ Particle. The resulting nucleus is
$_nX^m \rightarrow _{n-2}X^{m-4} \rightarrow _nY^{m-4}$
Af a certain time, a radio active sample contains $2 \times 10^{20}$ atoms and disintegration rate is $3 \times 10 ^ {10}$ atom persec. When$2 \times 10^{15}$ atoms are Left to decay its disintegation rate will be
${dN \over df} = df \Rightarrow I = dN$ $ I^1 = \lambda N_1 $ $ I_2 = \lambda N_2 $ $ {I_1 \over N_1} = \lambda $ $ {I_1 \over N_2} = \lambda $ $\therefore {I_1 \over N_1} = { f_2 \over N_2} \Rightarrow I_2 = { I_2 \over N_1} \times N_2 = { 3 \times 10^{10} \times 2 \times 10^{15} \over 2 \times 10_20} = 3 \times 10^5 $
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