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A particle having a speed of 0.87c has a momentum of 10-16 kg·m/s. what is its mass?

cupoosta [38]

The momentum p of a moving particle is the product between its mass, m, and tis velocity, v:
$p=mv$
In our problem, we know $p=10^{-16}~Kg \cdot m/s$ and $v=0.87c=0.87\cdot 3\cdot10^8~m/s=2.61\cdot 10^8~m/s$ , and using the relationship mentioned above, we can find the mass m of the particle:
$m= \frac{p}{v} =3.8\cdot10^{-25}~Kg$

8 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

4 years ago

A scientist is designing a device that will mimic Earth's atmosphere by

mafiozo [28]

Answer:

1. Ultraviolet light (UV)

2. X-rays

3. Gamma-rays

Explanation:

Though there are different types of energy or electromagnetic waves with varying wavelengths, including the likes of Gamma X-rays, ultraviolet light, visible light, infrared radiation, and microwave radiation.

What is more certain is that the atmosphere blocked the high-energy waves from getting to the earth surface or biosphere such as Ultraviolet light (UV), X-rays and Gamma-rays

8 0

3 years ago

Help ASAP plsssss:///

kenny6666 [7]

Answer:

c

Explanation:

8 0

3 years ago

Find the configuration of any tow

Yuki888 [10]

Answer:

<h2>Ok I choose Copper and Zinc , Here is your answer⤴️⤴️</h2><h3>Hope it's helpful for you mark me as brainlist please</h3>

6 0

3 years ago