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Sebuah partikel mempunyai energi relativistik total 10 GeV dan momentum relativistik 8 GeV/c. Massa diam partikel itu ialah....

Pertanyaan

Sebuah partikel mempunyai energi relativistik total 10 GeV dan momentum relativistik 8 GeV/c. Massa diam partikel itu ialah....space

  1. begin mathsize 14px style 0 comma 25 space GeV divided by straight c squared end style 

  2. begin mathsize 14px style 1 comma 20 space GeV divided by straight c squared end style 

  3. begin mathsize 14px style 2 comma 00 space GeV divided by straight c squared end style 

  4. begin mathsize 14px style 6 comma 00 space GeV divided by straight c squared end style 

  5. begin mathsize 14px style 16 comma 0 space GeV divided by straight c squared end style 

Pembahasan Soal:

Diketahui :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row E equals cell 10 space GeV end cell row p equals cell 8 space GeV divided by straight c end cell end table end style 

Ditanya :

begin mathsize 14px style m subscript o equals.... ? end style 

Jawab :

Rumus energi :

undefined 

Rumus momentum relativistik :

undefined 

Kita gunakan perbandingan E dan p untuk mencari kecepatan v :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E over p end cell equals cell c squared over v end cell row cell fraction numerator 10 space GeV over denominator 8 space GeV divided by c end fraction end cell equals cell c squared over v end cell row v equals cell 8 over 10 c end cell row v equals cell 0 comma 8 c end cell end table end style 

Kita cari nilai massa diamnya menggunakan nilai dari momentum relativistik :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row p equals cell m v end cell row p equals cell fraction numerator m subscript o v over denominator square root of 1 minus open parentheses begin display style v over c end style close parentheses squared end root end fraction end cell row cell 8 space G e V divided by c end cell equals cell fraction numerator m subscript o open parentheses 0 comma 8 c close parentheses over denominator square root of 1 minus open parentheses begin display style fraction numerator 0 comma 8 c over denominator c end fraction end style close parentheses squared end root end fraction end cell row cell 8 space G e V divided by c end cell equals cell fraction numerator m subscript o open parentheses o comma 8 c close parentheses over denominator 0 comma 6 end fraction end cell row cell m subscript o end cell equals cell 6 space GeV divided by straight c squared end cell end table end style 

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 13 Maret 2021

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