Question

calculate the minimum uncertainty in the velocity of an electron confined to a box of length 10^-10 given m- 9.1*10^31Kg h=6.63*10^34js

Answer 1

We have to calculate the minimum uncertainty in the velocity of an electron confined to a box of length 10¯¹⁰ m. mass of electron,m = 9.1 × 10¯³¹ kg and h = 6.63 × 10¯³⁴ Js

solution : we know, from Heisenberg's uncertainty principle,

∆v = h/4πm∆x

here ∆x = 10¯¹⁰ m , m = 9.1 × 10¯³¹ kg, h = 6.63 × 10¯³⁴ Js

∆v = (6.63 × 10¯³⁴)/(4 × 3.14 × 9.1 × 10¯³¹ × 10¯¹⁰)

= 0.058 × 10^7 m/s

= 5.8 ×10⁵ m/s

Therefore the minimum uncertainty in velocity is 5.8 × 10⁵ m/s.

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