QJEEAdv04Ch28e4: JEE Advanced 2004 Chemistry - Class 11 JEEAdv04Ch28e4 Solution: (a) The Schrodinger wave equation for hydrogen atom is Where a…

(a) Probability of finding electrons at any place = $\psi _{2s}^2 = 0$ at node

$\therefore$ $\psi _{2s}^2 = 0 = {1 \over {4\sqrt {2\pi } }}{\left( {{1 \over {{a_0}}}} \right)^3}\left( {2 - {r \over {{a_0}}}} \right) \times {e^{ - r/{a_0}}}$

or $\left( {2 - {r \over {{a_0}}}} \right) = 0$ or $2 = {r \over {{a_0}}}$ or $r = 2{a_0}$

(b) $\lambda = {h \over {mv}}$

$h = 6.626 \times {10^{ - 34}}\,Js$ , m = 100 g = 0.1 kg v = 100 ms ^{$-$ 1}

$\therefore$ $\lambda = {{6.626 \times {{10}^{ - 34}}\,kg\,{m^2}\,{s^{ - 1}}} \over {0.1\,kg \times 100\,m{s^{ - 1}}}}$

$= 6.626 \times {10^{ - 35}}\,m$