Question
The position of both, an electron and a helium atom is known within $$1.0\,nm.$$ 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.
$$8.0 \times {10^{ - 26}}kg\,m{s^{ - 1}}$$
D.
$$5.0 \times {10^{ - 26}}kg\,m{s^{ - 1}}$$
Answer :
$$5.0 \times {10^{ - 26}}kg\,m{s^{ - 1}}$$
Solution :
According to Heisenberg uncertainty principle,
$$\Delta x \times \Delta p = \frac{h}{{4\pi }}{\text{(which is constant)}}{\text{.}}$$
As $$\Delta x$$ for electron and helium atom is same, thus momentum of electron and helium will also be same, therefore the momentum of helium atom is equal to $$5.0 \times {10^{ - 26}}kg\,m{s^{ - 1}}.$$