\(\triangle\)x = 0.6\(\mathring{A}\) = 0.6 x 10^-10m
\(\triangle p\) = ?
\(\triangle x.\triangle p\ge\frac{h}{4\pi}\)
\(\triangle x.\triangle p\ge5.28\times10^{-35}kgm^{2}s^{-1}\)
\((0.6 \times10^{-10}) \triangle p \ge5.28\times10^{-35}\)
\(\Rightarrow \triangle \ge \frac{5.28\times10^{-35}kgm^{2}s^{-1}}{0.6510^{-1}m}\)
\(\triangle p \ge8.8\times10^{-25}kgms^{-1}\)

v = 2.85 x 10⁸ ms^-1
m_p = 1.673 x 10^-27Kg
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times10^{-34}kgm^{2}s^{-1}}{1.673\times10^{-27}kg\times\times2.85\times10^{8}ms^{-1}}\)
\(\lambda=1.389\times10^{-15}\Rightarrow\lambda=1.389\times10^{-15}A\) \([\because \mathring{A}={10}^{-10}m]\)

\({ He }^{ + }\longrightarrow { He }^{ 2+ }+{ e }^{ - }\)
\({ E }_{ n }=\frac{-13.6(2)^2}{n^2}\)
\({ E }_{ 1 }=\frac{-13.6(2)^2}{(1)^2}=-54.4\)
\({ E }_{ \infty }=\frac{-13.6(2)^2}{(\infty)^2}=0\)
\(\therefore\) Required Energy for the given process
= E_\(\infty\) - E_I = 0 - (- 54.4) = 56.4 ev.

Given that, y = x + 11.1% of x
\(=\left(x+\frac{11.1}{100} x\right)=x+0.111 x\)
Y = 1.111 x
mass number = 37
number of protons + number of neutrons = 37
(x -1) + 1.111x  = 37
x+1.111x = 38
2.111x  = 38
\(x=\frac{38}{2.11}\)
x = 18.009
x 18 (whole number)
ஃ  Atomic number = x- 1 = 18 - 1 = 17
Mass number = 37
Symbol of the ion \({ _{ 17 }^{ 37 }{ Cl } }^{ - }\)

Potential difference = 100V
= 100 x 1.6 x 10-¹⁹J
\(\lambda=\frac{h}{\sqrt{2mev}}\)
\(=\frac { 6.626\times { 10 }^{ -34 }Kg{ m }^{ 2 }{ s }^{ -1 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }Kg\times 100\times 1.6\times { 10 }^{ -19 }J } } \)
\(\lambda=1.22\times10^{-10}m\)