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Sebuah elektron bergerak dengan energi kinetik seb...

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 straight J times straight s 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)

  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 12 end exponent space straight m end style 

  4. begin mathsize 14px style 0 comma 721 cross times 10 to the power of negative 12 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 

Jawaban:

Louise de Broglie menyatakan pendapatnya bahwa cahaya dapat berkelakuan seperti partikel maka partikel pun seperti halnya elektron dapat berkelakuan seperti gelombang. Benda atau partikel yang bermassa m dan bergerak dengan kecepatan v akan memiliki momentum sebesar mv, sehingga panjang gelombang de Broglie dari benda partikel tersebut dinyatakan dengan persamaan:

begin mathsize 14px style lambda equals fraction numerator h over denominator m times v end fraction end style 

Pada soal diketahui, begin mathsize 14px style E K equals 2 over 3 E subscript 0 end stylebegin mathsize 14px style h equals 6 comma 626 cross times 10 to the power of negative 34 end exponent space straight J times straight s end stylem0 = begin mathsize 14px style 9 comma 1 cross times 10 to the power of negative 31 end exponent space kg end stylebegin 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, maka:

mencari kecepatan:

begin mathsize 14px style E K equals E minus E subscript 0 2 over 3 E subscript 0 equals E minus E subscript 0 5 over 3 E subscript 0 equals E 5 over 3 m subscript 0 c squared equals m c squared 5 over 3 m subscript 0 equals fraction numerator m subscript 0 over denominator square root of 1 minus open parentheses begin display style v over c end style close parentheses squared end root end fraction 5 square root of 1 minus open parentheses v over c close parentheses squared end root equals 3 25 open parentheses 1 minus open parentheses v over c close parentheses squared close parentheses equals 9 1 minus open parentheses v over c close parentheses squared equals 9 over 25 1 minus 9 over 25 equals open parentheses v over c close parentheses squared 16 over 25 equals open parentheses v over c close parentheses squared v over c equals square root of 16 over 25 end root v equals 0 comma 8 c end style 

massa:

begin mathsize 14px style m equals fraction numerator m subscript 0 over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction m equals fraction numerator m subscript 0 over denominator square root of 1 minus begin display style open parentheses 0 comma 8 c close parentheses squared over c squared end style end root end fraction m equals fraction numerator m subscript 0 over denominator square root of 1 minus 0 comma 64 end root end fraction m equals fraction numerator m subscript 0 over denominator 0 comma 6 end fraction m equals fraction numerator 9 comma 1 cross times 10 to the power of negative 31 end exponent over denominator 0 comma 6 end fraction equals 15 comma 17 cross times 10 to the power of negative 31 end exponent space kg end style 

Maka, panjang gelombang de Broglie:

begin mathsize 14px style lambda equals fraction numerator h over denominator m times v end fraction lambda equals fraction numerator 6 comma 626 cross times 10 to the power of negative 34 end exponent over denominator 15 comma 17 cross times 10 to the power of negative 31 end exponent open parentheses 0 comma 8 cross times 3 cross times 10 to the power of 8 close parentheses end fraction lambda equals fraction numerator 6 comma 626 cross times 10 to the power of negative 34 end exponent over denominator 36 comma 408 cross times 10 to the power of negative 23 end exponent end fraction lambda equals 0 comma 182 cross times 10 to the power of negative 11 end exponent space straight m lambda equals 1 comma 82 cross times 10 to the power of negative 12 end exponent space straight m end style 

Jadi, jawaban yang tepat adalah A.undefined 

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