Practice free Nuclei (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
It $N_o$ is the original mass of the substance of halt lift 5 years, the amount of substance left after 15 years is
${N \over no} = (1/2)^{t/T^{1/2}}$
when u-238 nucleus originally at lest decay by emitting an $\alpha$-particle having a speed u the recoil speed of the resi-dual nucleus is.
According to conservation of momentum
At a certain instant, a radioactive sample has a decay rate of 5000 dis-interation Per minute. After 5 minute the decay rate is 1250 dis-interations Per minute. Then the decay constant is (Per-min)
$N = N_oe^{-\lambda t} --> I = I_oe^{-\lambda t}$ $1250 = 5000 e^{-\lambda \times 5}$ ${1\over 4} = e^{-5 \lambda}$ $ 4 = e^{-5 \lambda }$ $ ln4 = 5 \lambda$ $$ \lambda = {1\over5 }ln4 = 0.2 ln4$$
A nucleus with Z=92 emits the following sequence $\alpha,\alpha,\beta^-,\beta^-,\alpha,\alpha,\alpha,\alpha,\beta^-,\beta^-,\alpha,\beta^+,\beta^+,\alpha$ . The Z of the resulting nucleus is
Atomic number of final nucleus = 92 -2 (no.of $\alpha$ - Particle) + 1 (No. of $\beta^-$
Particle)
It the radius of $_{13}^{27}Al$ nucleus is 3.6 fm the radius of $_{52}^{125}Te$ nucleus is nearly equal to
use formula $R= Ro (A)^{1/3}$
A radio-active nucleus $_Z^A{X}$ emits 3 $\alpha$ -particles and 2 Positrions. the ratio of number of neuleuons to that of Protons in the final nucleus will be
$\therefore$Number of neutrons= (A-12)-(Z-8) =A -12 - Z +8 =A - Z - 4 Number of proton = Z - 8 ${No.of neutrons \over No.of Protons} = {A-Z-4 \over Z-8} $
when $_3^7Li $ nuclear are bombarded by Proton and the resultant nuclei are 3 8 Be , the 4 emitted particle will be
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}} $
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 $
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