Question

Isolated neutrons outside the nucleus are unstable. After a typical lifetime of about 15 minutes, they decay into a proton and another particle called a neutrino. Consider a neutron that lives 900 s when at rest relative to an observer. How fast is the neutron moving relative to an observer who measures its life span to be 2265 s?

Answer 1

To solve the exercise it is necessary to apply the concepts of time dilation and the Lorentz factor, that is, the factor by which time, length, and relativistic mass change for an object while that object is moving.

The lorentz factor is defined as,

$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Where,

v = relative velocity between inertial reference frames

c = speed of light

On the other hand we have time dilation,

$\Delta t' = \gamma \Delta t$

Where $\Delta t'$ is he time between two ticks as measured in the frame in which the clock is moving and $\Delta t$ is the time between these ticks as measured in the rest frame of the clock.

$\gamma =$ Lorentz Factor

Applying this theory in our problem we have that,

$\Delta t' = \gamma \Delta t$

Re-arrange to Lorentz Factor,

$\gamma = \frac{\Delta t}{\Delta t'}$

$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{\Delta t}{\Delta t'}$

Replacing with our values,

$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{2265}{900}$

We need to arrange that expression to find the velocity, then

$\frac{1}{1-(\frac{v}{c})^2}=(\frac{2265}{900})^2$

$1-(\frac{v}{c})^2 = \frac{1}{(\frac{2265}{900})^2}$

$(\frac{v}{c})^2 = 1- \frac{1}{(\frac{2265}{900})^2}$

$v = c \sqrt{(1- \frac{1}{(\frac{2265}{900})^2})}$

$v = 0.91766c$

Therefore the velocity of the neutron is 0.91766 times the speed of light

Answer 2

The speed of the neuron moving relative to an observer will be $v=0.9176\frac{m}{s}$

What will be the speed of the neutron moving relative to an observer?

By applying the concept of Time dilation that is Lorentz equation

The Lorentz factor is defined as,

$\gamma =\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2} } }$

here,

v = relative velocity between inertial reference frames

c = speed of light

On the other hand, we have time dilation,

$\Delta t'=\gamma\Delta t$

Where is the time between two ticks as measured in the frame in which the clock is moving and is the time between these ticks as measured in the rest frame of the clock?

$\gamma=$ Lorentz Factor

Re-arrange to Lorentz Factor,

$\gamma=\dfrac{\Delta t}{\Delta t'}$

Replacing with our values,

$\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2} } }=\dfrac{\Delta t}{\Delta t'}$

We need to arrange that expression to find the velocity, then

$\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2} } }=\dfrac{2265}{900}$

$\dfrac{1}{{1-\dfrac{v^2}{c^2} } }=(\dfrac{2265}{900} )^2$

$(\dfrac{v}{c} )^2 =1-\dfrac{1}{(\dfrac{2265}{900})^2 }$

$v= 0.9176\frac{m}{s}$

Thus the speed of the neuron moving relative to an observer will be $v=0.9176\frac{m}{s}$

To know more about Time dilation follow

https://brainly.com/question/1933572