Given Data :- $ L = 10^{(-10)}m$

Formula :- $∆X_ma.∆p_mi= ħ$

Calculations :- since the electron is probable anywhere within the box of length $10^{(-10)}$m, the maximum uncertainty in locating it is,

$∆X_ma= 10^{(-10)}m$

$∆X_ma.m.∆v_mi= ħ$

$∆v_mi= ħ/(m∆X_ma )$

$ = (6.63×10^{(-34)})/(2×3.14×9.1×10^{(-31)}×10^{(-8)} )$

$∆v_mi=1.16×10^5m/sec$

Answer :- minimum uncertainty in velocity = $1.16×10^5$m/s