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natka813 [3]
2 years ago
13

Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez

ers is 1000 W/m2, the same as the intensity of sunlight at the surface of the Earth. (a) What is the pressure on an atom if light from the tweezers is totally absorbed? ? Pa (b) If this pressure were exerted on a tritium atom, what would be its acceleration? (The mass of a tritium atom is 5.01 ✕ 10−27 kg. Assume the cross-sectional area of the laser beam is 6.65 ✕ 10−29 m2.)
Physics
1 answer:
Zanzabum2 years ago
6 0

(a)  3.3\cdot 10^{-6} Pa

The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

p=\frac{I}{c}

where

I is the intensity of the wave

c is the speed of light

In this problem,

I=1000 W/m^2

and substituting c=3\cdot 10^8 m/s, we find the radiation pressure

p=\frac{1000 W/m^2}{3\cdot 10^8 m/s}=3.3\cdot 10^{-6}Pa

(b) 4.4\cdot 10^{-8} m/s^2

Since we know the cross-sectional area of the laser beam:

A=6.65\cdot 10^{-29}m^2

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

F=pa=(3.3\cdot 10^{-6}Pa)(6.65\cdot 10^{-29} m^2)=2.2\cdot 10^{-34}N

And then, since we know the mass of the atom

m=5.01\cdot 10^{-27}kg

we can find the acceleration, by using Newton's second law:

a=\frac{F}{m}=\frac{2.2\cdot 10^{-34} N}{5.01\cdot 10^{-27} kg}=4.4\cdot 10^{-8} m/s^2

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Answer:

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Explanation:

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The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}

By the law of conservation we know that P_{i} =P_{f}

For P_{f} (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

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P_{i} =P_{f}=112.5\\

112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}

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Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

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The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

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Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}

Put the value into the formula

Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}

Q=47.8\times10^{-9}\ C

Q=47.8\ nC

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