Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

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Explanation- dN/dt= $\lambda N$

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Explanation- nuclei ₂He⁴ and ₁He³ have the same mass number . the element having more number of proton having more repusion so less binding energy. So ₂He⁴ have more number of proton so more repulsion so less binding energy.

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Explanation- B.E = (M_p+M_H-M_N)c²

B.E= (118.9058+1.0078252-119.902199)c²

B.E=0.0114362 c²

B.E= (M_p+M_H-M_N)c²

B.E= (119.902199+1.0078252-120.902822)c²

B.E= 0.0059912c²

  (ii) the existence of magic numbers indicates that the shell structure of nucleus is similar to the shell structure of an atom. This also explains peaks in binding energy curve.

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Explanation- 

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Explanation- P_n=P_p+P_e

P_p+P_e=0

P_e=P_p=P

E_p= ${(m}_p^2{c}^4+p_p^2{c}^2)$ ^1/2

E_e= ${(m}_e^2{c}^4+p_p^2{c}^2)$ ^1/2

= ${(m}_e^2{c}^4+p_e^2{c}^2)$^1/2

From conservation of energy

^{${(m}_p^2{c}^4+p_p^2{c}^2)$ 1/2}= ${(m}_e^2{c}^4+p_e^2{c}^2)$ ^1/2= m_nc²

m_pc²= 936MeV, m_nc²=938MeV, m_ec²=0.51MeV

since the energy difference n and p is small

m_pc²+ $\frac{p^2{c}^2}{2m_p^2{c}^4}=m_n{c}^2-pc$

pc= m_nc²-m_pc² = 938-936= 2MeV

E_p=( $m_p^2{c}^4+p^2{c}^2$ )^1/2= $\sqrt[]{{0.51}^2+2^2}$

= 2.06MeV

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Explanation-E= $\frac{m{e}^4}{\pi {\epsilon }_0^2{h}^2}=13.6eV$

If proton and neutron had charge e each and were governed by the same electrostatic force, then in the above equation we would need relace m

So m' = $\frac{M*N}{M+N}$ =M/2=918m

Hence binding energy= 918me'⁴/8 ${\epsilon }_0^2{h}^2=2.2MeV$

Dividing eqn

918 ( $\frac{e\text{'}}{e})$ ⁴= $\frac{2.2MeV}{13.6eV}$ = $\frac{2.2*{10}^6}{13.6}$

$\frac{e\text{'}}{e}$= 3.64

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Explanation- B.E= 2.2MeV

E-B=K_n+K_P= $\frac{p_n^2}{2m}+\frac{p_p^2}{2m}$

Conservation of momentum= P_n+P_P= E/C

E=B $p_n^2-p_p^2=0$

It only happen if P_n=P_p=0

Let E=B+X where X<

X= (E/C-P_P)/2m+ $\frac{P_0^2}{2m}$

2 $P_P^2-\frac{2E{P}_p}{c}$ + $\frac{E^2}{C^2}-2mX$ =0

Pp= $\frac{\frac{2E}{c}?\text{exception 25:}}{4}$

$\frac{4E^2}{C^2}=8(\frac{E^2}{c^2}-2mX)$

$16mX$ = 4E²/c²

$X=\frac{E^2}{4m{c}^2}=\frac{B^2}{4m{C}^2}$

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Explanation- let ³⁸S have N₁ active nuclei and ³⁸Cl have N₂ active nuclei

$\frac{d{N}_1}{dt}={-\lambda }_1{N}_1$ + ${\lambda }_1{N}_1$

N₁=N_{0 $e^{-\lambda 1t}$}

$\frac{d{N}_2}{dt}={-\lambda }_1{N}_2{e}^{\left({-\lambda }_{1}t\right)}$ + ${\lambda }_2{N}_2$

Multiplying by $e^{\lambda 2t}dt$ and then integrating both sides we got

N2= $\frac{N_o{\lambda }_1}{{\lambda }_2-{\lambda }_1}(e^{-\lambda 1t}-e^{-\lambda 2t})$

After solving it we get time t= (log $\frac{{\lambda }_1}{{\lambda }_2}$ )/ ${\lambda }_1-{\lambda }_2$

t= $\frac{log\frac{2.48}{0.62}}{2.48-0.62}$ = $\frac{2.303*2*0.3010}{1.86}=0.745s$