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3 years ago

What happens when a proton is brought near another proton?

Physics

1 answer:

muminat3 years ago

5 0

They will repel each other because they both have positive charges.

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The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

7 0

3 years ago

Which formula is used to calculate average velocity?

egoroff_w [7]

Average velocity is the change in distance divided by the time it takes to travel that distance. so the first one.

7 0

3 years ago

Read 2 more answers

A disk between vertebrae in the spine is subjected to a shearing force of 425 N. Find its shear deformation, taking it to have a

zzz [600]

Answer:

option A

Explanation:

given,

shear force = 425 N

Shear modulus = 1.7 × 10⁹ N/m²

disk equivalent to solid cylinder

height = 0.7 cm and diameter = 6.50 cm

$\delta = \dfrac{VL}{AG}\\\\\delta = \dfrac{425\times 0.007}{0.25 \times \pi d^2\times 1.7\times 10^9}\\\\\delta = \dfrac{425\times 0.007}{0.25 \times \pi\times 0.065^2\times 1.7\times 10^9}\\\\\delta = 5.27 \times 10^{-7} m$