Class 12th

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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

$\begin{array}{l}{\displaystyle {\int }_0^{1}x{e}^{x^2}dx=Isay},\\ Putting,x^2=t\Rightarrow 2xdx=dt\to xdx=\frac{dt}{2}\\ ={\displaystyle {\int }_0^1{e}^t\frac{dt}{2}}=\frac{1}{2}{\displaystyle {\int }_0^1{e}^{t}dt}\end{array}$

When,

x = 0, t = 0

x = 1, t = 1

$\begin{array}{l}=\frac{1}{2}{\left[e^t\right]}_0^1\\ =\frac{1}{2}\left[e^1-e^0\right]=\frac{\left(e-1\right)}{2}\end{array}$

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

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