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3 years ago

Which is the best predictor of the radioactive nature of an isotope?

Physics

2 answers:

frutty [35]3 years ago

8 0

Answer:

B. The neutron-to-proton ratio

Explanation:

ale4655 [162]3 years ago

7 0

The answer is <span>the neutron-to-proton ratio</span>

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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

3 years ago

A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

muminat

Answer:

The speed of the plank relative to the ice is:

$v_{p}=-0.33\: m/s$

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

$m_{g}v_{g}+m_{p}v_{p}=0$ (1)

Where:

m(g) is the mass of the girl
m(p) is the mass of the plank
v(g) is the speed of the girl
v(p) is the speed of the plank

Now, as we have relative velocities, we have:

$v_{g/b}=v_{g}-v_{p}=1.55 \: m/s$ (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

$45v_{g}+168v_{p}=0$

$v_{g}-v_{p}=1.55$

$v_{p}=-0.33\: m/s$

I hope it helps you!

8 0

2 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi

elena55 [62]

The angular acceleration of a rotating object is given by
$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$
where
$\omega_f$ is the final angular speed of the object
$\omega_i$ is its initial angular speed
$\Delta t$ is the time taken to accelerate

For the wheel in our problem, $\omega_f=11.1 rad/s$ , $\omega_i = 0$ and $\Delta t=2.99 s$ , so its angular acceleration is
$\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2$

8 0

2 years ago

Name two things that the people standing on the street corner would say are in motion

timurjin [86]

People, and cars
is the right answer!!!!!!!!!!!!!!!!!!!!!!!!!!

8 0

2 years ago

Which substance is a heterogeneous mixture?

anyanavicka [17]

Answer:

A heterogeneous mixture is simply any mixture that is not uniform in composition - it's a non-uniform mixture of smaller constituent parts.

Explanation: