$(a)\;0.70nm\qquad(b)\;0.60nm\qquad(c)\;0.40nm\qquad(d)\;0.80nm$

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V=$4.5\times10^5$m/sec

$\bigtriangleup V= V\times \large\frac{0.01}{100}$

$=4.5\times10$

=45m/sec

$\bigtriangleup x .\bigtriangleup V \approx \large\frac{h}{4\pi m}$

$\bigtriangleup x = \large\frac{h}{4\pi m\bigtriangleup V}$

$=\large\frac{6.626\times10^{-34}}{4\pi\times1.673\times10^{-27}\times27}$

$=7\times10^{-10}$

=0.70nm

Hence the anwer is (a)

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