MHTCET 17 Apr 2026 Shift 1 Paper

Concept:Heisenberg's uncertainty principle states that the product of uncertainties in position and momentum of a particle is at least $\frac{h}{4\pi}$.Explanation:The uncertainty principle is expressed as: $\Delta x \Delta p \geq \frac{h}{4\pi}$.Since momentum $p = m_e v$, the uncertainty in momentum is $\Delta p = m_e \Delta v$.Rearranging the equation gives the uncertainty in velocity: $\Delta v = \frac{h}{4\pi m_e \Delta x}$.Convert the given position uncertainty into metres: $\Delta x = 50 \text{ pm} = 50 \times 10^{-12} \text{ m}$.Substitute the given values into the formula:$$\Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.142 \times 9.1 \times 10^{-31} \times 50 \times 10^{-12}}$$Simplifying the denominator: $4 \times 3.142 \times 9.1 \times 10^{-31} \times 50 \times 10^{-12} = 5.72 \times 10^{-40}$.Therefore, $\Delta v = \frac{6.63 \times 10^{-34}}{5.72 \times 10^{-40}} \approx 1.16 \times 10^6 \text{ m s}^{-1}$.Answer:The uncertainty in the velocity of the electron is $1.16 \times 10^6 \text{ m s}^{-1}$.Hence, the correct option is B.