Question

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c

Answer 1

Answer:

0.85c

Explanation:

Rest mass of Kaon $M_{0K}$ = 494 MeV/c²

Rest mass of proton $M_{0P}$  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy $E_{0K}$ = 494c² MeV

for the proton, rest energy $E_{0P}$ = 938c² MeV

Recall that the rest energy, and the total energy are related by..

$E$ = γ$E_{0}$

which can be written in this case as

$E_{K}$ = γ$E_{0K}$ ...... equ 1

where $E$ = total energy of the kaon, and

$E_{0}$ = rest energy of the kaon

γ = relativistic factor = $\frac{1}{\sqrt{1 - \beta ^{2} } }$

where $\beta = \frac{v}{c}$

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

$E_{K}$ = $E_{0P}$ ......equ 2

where $E_{K}$ is the total energy of the kaon, and

$E_{0P}$ is the rest energy of the proton.

From $E_{K}$ = $E_{0P}$ = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = $\frac{1}{\sqrt{1 - \beta ^{2} } }$ = 1.89

1.89$\sqrt{1 - \beta ^{2} }$ = 1

squaring both sides, we get

3.57( 1 - $\beta^{2}$) = 1

3.57 - 3.57$\beta^{2}$ = 1

2.57 = 3.57$\beta^{2}$

$\beta^{2}$ = 2.57/3.57 = 0.72

$\beta = \sqrt{0.72}$ = 0.85

but, $\beta = \frac{v}{c}$

v/c = 0.85

v = 0.85c