Roboguru

Sebuah proton massa diamnya  bergerak dengan energi kinetik sebesar 10 GeV. Hitunglah laju proton!

Pertanyaan

Sebuah proton massa diamnya begin mathsize 14px style 1 , 67 cross times 10 to the power of negative 27 end exponent blank kg end style bergerak dengan energi kinetik sebesar 10 GeV. Hitunglah laju proton!

Pembahasan Soal:

Gunakan perumusan energi kinetik.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E K end cell equals cell left parenthesis gamma minus 1 right parenthesis E subscript 0 end cell row gamma equals cell 1 plus fraction numerator E K over denominator E subscript 0 end fraction end cell row cell fraction numerator 1 over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction end cell equals cell 1 plus fraction numerator E K over denominator E subscript 0 end fraction end cell row cell square root of 1 minus v squared over c squared end root end cell equals cell fraction numerator 1 over denominator 1 plus begin display style fraction numerator E K over denominator E subscript 0 end fraction end style end fraction end cell row cell 1 minus v squared over c squared end cell equals cell open parentheses fraction numerator 1 over denominator 1 plus begin display style fraction numerator E K over denominator E subscript 0 end fraction end style end fraction close parentheses squared end cell row cell v squared over c squared end cell equals cell 1 minus open parentheses fraction numerator 1 over denominator 1 plus begin display style fraction numerator E K over denominator E subscript 0 end fraction end style end fraction close parentheses squared end cell row cell v over c end cell equals cell square root of 1 minus open parentheses fraction numerator 1 over denominator 1 plus begin display style fraction numerator E K over denominator E subscript 0 end fraction end style end fraction close parentheses squared end root end cell row v equals cell c square root of 1 minus open parentheses fraction numerator 1 over denominator 1 plus begin display style fraction numerator E K over denominator E subscript 0 end fraction end style end fraction close parentheses squared end root end cell end table end style  

Selanjutnya, kita akan mencari energi diam proton melalui persamaan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E subscript 0 end cell equals cell m subscript 0 c squared end cell row blank equals cell 1 comma 67 cross times 10 to the power of negative 27 end exponent space kg left parenthesis 3 cross times 10 to the power of 8 space straight m divided by straight s right parenthesis squared equals 15 comma 03 cross times 10 to the power of negative 11 end exponent space straight J end cell row blank equals cell 939 space MeV end cell row blank equals cell 0 comma 939 space GeV end cell end table end style 

Dengan demikian, 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell c square root of 1 minus open parentheses fraction numerator 1 over denominator 1 plus begin display style fraction numerator E K over denominator E subscript 0 end fraction end style end fraction close parentheses squared end root end cell row blank equals cell c square root of 1 minus open parentheses fraction numerator 1 over denominator 1 plus begin display style fraction numerator 10 space GeV over denominator 0 comma 939 space GeV end fraction end style end fraction close parentheses squared end root end cell row blank equals cell c square root of 1 minus open parentheses fraction numerator 1 over denominator 11.65 end fraction close parentheses squared end root end cell row blank equals cell c square root of 1 minus open parentheses fraction numerator 1 over denominator 11.65 end fraction close parentheses squared end root end cell row blank equals cell c square root of 1 minus open parentheses 0 comma 086 close parentheses squared end root end cell row blank equals cell c square root of 1 minus 0 comma 007 end root end cell row blank equals cell c square root of 0 comma 993 end root end cell row blank equals cell 0 comma 996 c end cell end table end style 

Dengan demikian, laju proton adalah 0,996c.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 15 Maret 2021

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