What is the total distance between the diving board and the diver's stopping point underwater?

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the 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, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

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

here, $\vec{P_\lambda}$ 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,

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

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

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

8 0

3 years ago

Find the angle of refraction of a ray of light that enters a diamond (n=2.419) from air at an angle of 18.0° to the normal. ONLY

Tpy6a [65]

Answer:

n

Explanation:

hb

3 0

3 years ago

A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy. The energy used to do work

nika2105 [10]

Answer:

480J

Explanation:

Using the formula:

Delta U = Q - W

Q:Heat (J)

Delta U: Changes in internal Energy (J)

W:Work (J)

We can plug in the give numbers, Q and W.

Delta U = 658J - 178J = 480J

6 0

2 years ago

2. A cinder block is sitting on a platform 20 m high. It weighs 16kg. The block has

Cerrena [4.2K]

Answer:

3136 Joules

Explanation:

Applying,

P.E = mgh.............. Equation 1

Where P.E = potential energy, m = mass of the cinder block, h = height of the platform, g = acceleration due to gravity.

From the question,

Given: m = 16 kg, h = 20 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 16(20)(9.8)

P.E = 3136 Joules

Hence the potential energy of the cinder block is 3136 Joules

7 0

2 years ago

If a star is found directly above the sun on the h-r diagram, what can you conclude about its size?.

hammer [34]

Answer:

It is larger than the Sun.

Explanation:

Brainliest pls

:3

7 0

1 year ago