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Sebuah elektron bergerak dengan energi kinetik sebesar 2/3 energi diamnya. Berapakah panjang gelombang de Broglie elektron tersebut? (Gunakan konstanta Planck , massa diam elektron , dan )

Pertanyaan

Sebuah elektron bergerak dengan energi kinetik sebesar 2/3 energi diamnya. Berapakah panjang gelombang de Broglie elektron tersebut? (Gunakan konstanta Planck begin mathsize 14px style h equals 6 comma 626 cross times 10 to the power of negative 34 end exponent space Js end style, massa diam elektron begin mathsize 14px style 9 comma 1 cross times 10 to the power of negative 31 end exponent space kg end style, dan begin mathsize 14px style 1 space eV equals 1 comma 6 cross times 10 to the power of negative 19 end exponent space straight J end style)undefined

  1. begin mathsize 14px style 1 comma 82 cross times 10 to the power of negative 12 end exponent space straight m end style 

  2. begin mathsize 14px style 1 comma 361 cross times 10 to the power of negative 12 end exponent space straight m end style 

  3. begin mathsize 14px style 1 comma 261 cross times 10 to the power of negative 21 end exponent space straight m end style 

  4. begin mathsize 14px style 0 comma 721 cross times 10 to the power of negative 21 end exponent space straight m end style 

  5. begin mathsize 14px style 0 comma 626 cross times 10 to the power of negative 12 end exponent space straight m end style 

Pembahasan Soal:

Diketahui :

begin mathsize 14px style E K equals 2 over 3 E subscript o end style 

Ditanya :

begin mathsize 14px style lambda equals.... ? end style 

Jawab :

Kita cari energi totalnya dulu :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E K end cell equals cell E minus E subscript o end cell row cell 2 over 3 E subscript o end cell equals cell E minus E subscript o end cell row E equals cell 5 over 3 E subscript o end cell end table end style 

Kita gunakan hubungan energi dan momentum relativistik untuk mencari momentum :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E squared end cell equals cell E subscript o superscript 2 plus open parentheses p c close parentheses squared end cell row cell open parentheses 5 over 3 E subscript o close parentheses squared end cell equals cell E subscript o superscript 2 plus open parentheses p c close parentheses squared end cell row cell 25 over 9 E subscript o superscript 2 end cell equals cell E subscript o superscript 2 plus open parentheses p c close parentheses squared end cell row cell 25 over 9 E subscript o superscript 2 minus E subscript o superscript 2 end cell equals cell open parentheses p c close parentheses squared end cell row cell open parentheses p c close parentheses squared end cell equals cell 16 over 9 E subscript o superscript 2 end cell row cell p c end cell equals cell 4 over 3 E subscript o end cell row cell p c end cell equals cell 4 over 3 m subscript o c squared end cell row p equals cell 4 over 3 m subscript o c end cell end table end style 

Rumus panjang gelombang de Broglie :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row p equals cell h over lambda end cell row lambda equals cell h over p end cell row lambda equals cell fraction numerator h over denominator begin display style 4 over 3 end style m subscript o c end fraction end cell row lambda equals cell fraction numerator 6 comma 626 cross times 10 to the power of negative 34 end exponent over denominator begin display style 4 over 3 end style cross times 9 comma 1 cross times 10 to the power of negative 31 end exponent cross times 3 cross times 10 to the power of 8 end fraction end cell row lambda equals cell 1 comma 82 cross times 10 to the power of negative 12 end exponent space straight m end cell end table end style 

Jadi, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 13 Maret 2021

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