Question

what will be energy of proton if it is moving with 97% speed of light?

Answer 1

Answer:

Explanation:

HERE THE ENERGY REFERS TO KINETIC ENERGY SINCE THE PROTON IS MOVING

VELOCITY , V = 3*10⁸ M/S

MASS = 1.6*10⁻²⁷KG

KINETIC ENERGY = 0.5 M V²

GIVEN 97 % OF ENERGY POSSESED BY PROTON

ENERGY = 97/100*0.5*1.6*10⁻27*3*10⁸

              = (*97*0.5*1.6*3)*10⁻²¹

              = 232.8*10⁻²¹ KG M²/S².

HOPE THIS HELPS.

Answer 2

Answer:

E = 6.1922224e-10 Joules

Explanation:

This can be done by applying principles of relativity, as we are approaching speeds close to c.

The energy (i.e. the total energy) E can be given by:

E = Kinetic Energy + Rest mass (mc^2)

$E= K +mc^2\\=\gamma mc^2$

Here $\gamma$ is the Lorentz factor. This is applied to adjust the dilations in time and length occurring at speeds close to c. You cannot simply apply Gallelian Transforms here.(i.e. A point (x,y,z) in a frame of reference S at rest is not (x+ut,y,z) in a frame of ref. S' moving with speed u on x-axis.)

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

So here if you are wondering where the factor got introduced in K + mc^2,

We use the work–energy theorem: $K=W=\int_{x_1}^{x_2}Fdx\\=\int_0^v\frac{mv_x}{(\gamma_x)^{\frac{-3}{1}}}\,dv_x$

By simple change of variables the integral is evaluated to $K=\gamma mc^2-mc^2=(\gamma -1)mc^2$

Add it with the rest energy mc^2 to get the value of $E=\gamma mc^2$

So here,

$E= \gamma mc^2=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}mc^2=\frac{1}{\sqrt{1-(0.97)^2}} \times 1.6726219 \times 10^{-27} \times (3 \times10^{8})^2$

E = 6.1922224e-10 Joules

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