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

0.85c

Step-by-step 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 = (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 = √(0.72)` = 0.85

but,
`\beta = (v)/(c)`

v/c = 0.85

v = 0.85c