Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter $10^{-15} \mathrm{~m}$ is $\_\_\_\_$ $\times 10^{9} \mathrm{~ms}^{-1}$ (nearest integer) [Given : mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$, Plank's constant $(\mathrm{h})=6.626 \times 10^{-34} \mathrm{Js}$] (Value of $\pi=3.14$)

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Atomic Structure

Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter

1 0^{- 15} m

____

\times 1 0^{9} ms^{- 1}

(nearest integer)
[Given : mass of electron

= 9.1 \times 1 0^{- 31} kg

, Plank's constant

(h) = 6.626 \times 1 0^{- 34} Js

]
(Value of

π = 3.14

)

Answer: 58

Heisenberg's uncertainty principle:

Δ x \cdot m Δ v \geq \frac{h}{4 π}

. Taking the minimum uncertainty:

Δ v = \frac{h}{4 π m Δ x} = \frac{6.626 \times 1 0 ^{- 34}}{4 \times 3.14 \times 9.1 \times 1 0 ^{- 31} \times 1 0 ^{- 15}}

= \frac{6.626 \times 1 0 ^{- 34}}{1.1424 \times 1 0 ^{- 44}} \approx 57.97 \times 1 0^{9} \approx 58 \times 1 0^{9} ms^{- 1}

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Integer answer