Question
The position of both, an electron and a helium atom is known within $$1.0\,mm.$$ Further the momentum of the electron is known within $$5.0 \times {10^{ - 26}}kg\,m{s^{ - 1}}.$$ The minimum uncertainty in the measurement of the momentum of the helium atom is
A.
$$50\,kg\,m{s^{ - 1}}$$
B.
$$80\,kg\,m{s^{ - 1}}$$
C.
$$80 \times {10^{ - 26}}kg\,m{s^{ - 1}}$$
D.
$$5.0 \times {10^{ - 26}}kg\,m{s^{ - 1}}$$
Answer :
$$5.0 \times {10^{ - 26}}kg\,m{s^{ - 1}}$$
Solution :
By Heisenberg’s uncertainty principle
$$\Delta x \times \Delta p \geqslant \frac{h}{{4\pi }}$$
when the position of electron and helium atom is same and momentum of electron is known within a range, therefore the momentum of helium atom is also equal to the momentum of electron, i.e.
$$5 \times {10^{ - 26}}kg\,m\,{s^{ - 1}}$$