$(a)\;1.05\times10^{-10}m\qquad(b)\;2.3\times10^{-13}m\qquad(c)\;1.3\times10^{-9}m\qquad(d)\;1.05\times10^{-13}m$

Velocity = $\large\frac{3\times10^8}{10}$

$=3\times10^7$m/sec

$\bigtriangleup V = V\times \large\frac{1}{100} = 3\times 10^5$

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

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

$=\large\frac{6.626\times10^{-34}}{4\pi\times1.673\times10^{-27}\times3\times10^5}$

$=1.05\times10^{-13}$m

Hence answer is (d)

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