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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)
The graph of displacement v/s time is
Its corresponding velocity-time graph will be
S vs.t is parabola
So V vs. t must be line and at t = 0 ds/dt>0
So option 1 is correct graph.
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 $R= Ro (A)^{1/3}$
If the binding energy of electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron form the first state of $Li^{2+}$ is.
Binding energy = ${13.6 Z^2 \over n^2}$ For $Li^{2+}$ Z = 3, n = 2 first exited state
The ionization Potential of hydrogen atom is 13.6 eV. An electron in the ground state absorbs Photon of energy 12.75 eV. How many different spectral lines can one expect when electron make a down ward transition
$\triangle E = E_n - E_1$ $E_n = \triangle E + E_1 = 12.75-13.6 = -0.85$ $E_n = -{13.6 \over n^2}$ $ -0.85 = {-13.6 \over n^2} $ $ n^2 = {13.6 \over 0.85} = 16 --> n=4 $
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} $
An $\alpha$ -particle of energy ${1\over2} mv^2$ bombards by a heavy nuclear target ofcharge ze.Then the distance of closet approach for the alpha nucleus will be Proportional to
at distance of closest approach K.E = P.E ${1\over 2}mv^2 = {1 \over 4 \pi \epsilon_o} {(ze) (ze) \over ro} \Rightarrow ro = {ze^2 \over \pi epsilon_o mv^2 }$
when $_3^7Li $ nuclear are bombarded by Proton and the resultant nuclei are 3 8 Be , the 4 emitted particle will be
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