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Starting with a sample of Puer cu-66, $7 \over 8$ of it decays into Zn, 15 minules the left of the sample is
$ {N \over N_o} = ({1 \over 2}) where \, n = {t \over T_{1\over 2}} $
An $\alpha $ -particle of energy 5 MeV is scattered though 180 by a fixed uranium nucleus. The distance of the closest approach nucleus The distance of the closest approach is of the order of
The distance of closest approach in a Rutherford scattering experiment can be estimated using the formula:
$$ d = rac{1}{4 \\pi \\epsilon_0} \\frac{2Ze^2}{E} $$
where $Z$ is the atomic number of the uranium nucleus (92), $e$ is the elementary charge, and $E$ is the energy of the alpha particle (5 MeV). Substituting these values, the order of magnitude for the distance of closest approach comes out to be approximately $10^{-12}$ cm.
The binding energy Per nucleon of deutron $(_1^2H)$ and Lielium nucleus $(_2^4{He})$ is 1.1 MeV and 7.0 MeV.respectively. If two neutron nuclear react to form a single helium nucleus, the energy released is
$_1H^2 + _1H^2 \rightarrow _2He^4$ $ \therefore B.E , of Helium = 4(7)-[2(1.1)+2(1.1)] = 28-4.4 = 23.6 eV$ $\therefore energy \, relewed \,is\, 23.6 eV $
The nucleus at rest disintegrate into two nuclear parts which have their velocities in the ratio 2:1 The ratio of their nuclear sizes will be
According to conservation momentum $m_1 v_1 = m_2 v_2 $ $ {m_1 \over m_2} = |{v_2 \over v1}| ={1\over 2} $ ${r_1^3 \over r_2^3} ={1 \over 2} $ $ {r_1 \over r_2} = ({1\over3})^{1 \over3}$ $r1:r2 = 1 : 3 \sqrt 2 $
A radiation of energy E falls normally on a Perfect reflecting surface. The momentum transferred to the surface is.
Here, surface is perfacet reflector momentum of incident radiation is E/C momentum of reflected rediation is - E/C change in momentum = - Momentum transtered to the surface = $2E \over C$
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 }$
which of the following transition in hydrogen atoms emits Photon of highest frequency?
n = 2 to n = 6 absorbs photon n = 9 to n = 2 absorbs photon n = 6 to n = 2 emission of photon n = 2 to n = 1 emission of photon $E = E_2-E_1 = {-13.6 \over 36}-({-13.6 \over 4 }) = -0.38+13.6 = 3.02 eV $ $ \triangle \Sigma^1 = E_2-E_1 = {-13.6 \over 4}-({-13.6 \over 1}) = -3.4+13.6 = 10.2 eV $ $ \triangle \Sigma^1 > \triangle E$ $hf' > hf $
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