# A microscope using suitable photon is employed to 0.1 A° . What is the uncertainty involved in the measurement of its velocity

jyoti bhatia
202 Points
7 years ago
According to Heisenberg’s Uncertainty principle:$\Delta x\Delta p> h/2\pi$
m = 9.11 x 10^-31 kg
h = 6.626 x 10^-34 Js
Therefore,
$\Delta v\geq h/4\pi m\Delta x$

jyoti bhatia
202 Points
7 years ago
According to Heisenberg’s Uncertainty principle:
$\Delta x\Delta p> h/2\pi$
m = 9.11*10exp-31 kg
h = 6.626*10exp-34Js
Therefore,
$\Delta v\geq h/4\pi m\Delta x$
$\Delta v\geq$6.626*10exp-34/4*3.14*9.11*10^-31*0.1*10^-10
$\Delta v\geq$0.579*10exp7 m/s or 5.79*10exp6 m/s

13 Points
6 years ago
According to the uncertainity principle of Heisenberg:
We can say that,
∆x(Position)*∆p(Momentum)=h/4π
Also,according to the specified question we have given that:
∆x=0.1Å=0.1×10^-10 metres or10^-11
h=6.6×10^-34
m=9.11×10^-31
We know that ∆p=m×∆v
So we can write ∆p in this form as we have to find the velocity
Now,Put these values up in the formula:
∆x*m∆v=h/4π
∆v=h/4π×∆x*m     (Take π=22/7)
∆v=6.6×10^-34×7/4×22×10^-11×9.11×10^-31
∆v=3.52×10^-24 m s^-1
or m/s