Class 12th

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This is a Short Answer Type Questions as classified in NCERT Exemplar

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.

$\begin{array}{l}{\displaystyle {\int }_0^2\frac{6x+3}{x^2+4}}dx={\displaystyle {\int }_0^2\frac{6x}{x^2+4}}dx+{\displaystyle {\int }_0^2\frac{3}{x^2+4}}dx\\ =3{\displaystyle {\int }_0^2\frac{2x}{x^2+4}}dx+3{\displaystyle {\int }_0^2\frac{1}{x^2+2^2}}dx\\ =3{\left[log\left|x^2+4\right|\right]}_0^2+\frac{3}{2}{\left[ta{n}^{-1}\frac{x}{2}\right]}_0^2\\ =3\left[log\left|2^2+4\right|-log\left|0^2+4\right|\right]+\frac{3}{2}\left[ta{n}^{-1}\frac{2}{2}-ta{n}^{-1}\frac{0}{2}\right]\\ =3\left[log\frac{8}{4}\right]+\frac{3}{2}[\pi /4-0]\\ =3log2+\frac{\mathrm{3\pi }}{}\end{array}$

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This is a Long Answer Type Questions as classified in NCERT Exemplar

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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This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- 

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This is a Long Answer Type Questions as classified in NCERT Exemplar

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