Roboguru

Sebuah foton yang panjang gelombangnya 0,070 nm mengalami hamburan Compton dengan sudut 53° (sin 53°= 0,8). Berapa besar energi yang diberikan kepada elektron yang terpental?

Pertanyaan

Sebuah foton yang panjang gelombangnya 0,070 nm mengalami hamburan Compton dengan sudut 53° (sin 53°= 0,8). Berapa besar energi yang diberikan kepada elektron yang terpental?undefined 

Pembahasan Soal:

Panjang gelombang foton setelah terhambur

begin mathsize 14px style lambda apostrophe minus lambda equals fraction numerator h over denominator m subscript e c end fraction left parenthesis 1 minus cos theta right parenthesis lambda apostrophe minus left parenthesis 0 comma 07 cross times 10 to the power of negative 9 end exponent right parenthesis equals fraction numerator 6 comma 63 cross times 10 to the power of negative 34 end exponent over denominator left parenthesis 9 comma 1 cross times 10 to the power of negative 31 end exponent right parenthesis left parenthesis 3 cross times 10 to the power of 8 right parenthesis end fraction left parenthesis 1 minus cos 53 right parenthesis lambda apostrophe minus left parenthesis 0 comma 07 cross times 10 to the power of negative 9 end exponent right parenthesis equals 0 comma 097 cross times 10 to the power of negative 11 end exponent lambda apostrophe equals 0 comma 097 cross times 10 to the power of negative 11 end exponent plus 7 cross times 10 to the power of negative 11 end exponent lambda apostrophe equals 7 comma 097 cross times 10 to the power of negative 11 end exponent space straight m end style 

Energi foton setelah hamburan
begin mathsize 14px style E apostrophe equals fraction numerator h c over denominator lambda apostrophe end fraction E apostrophe equals fraction numerator left parenthesis 6 comma 63 cross times 10 to the power of negative 34 end exponent right parenthesis left parenthesis 3 cross times 10 to the power of 8 right parenthesis over denominator left parenthesis 7 comma 097 cross times 10 to the power of negative 11 end exponent right parenthesis end fraction E apostrophe equals 2 comma 79 cross times 10 to the power of negative 15 end exponent space straight J end style  

Energi foton sebelum hamburan
begin mathsize 14px style E equals fraction numerator h c over denominator lambda end fraction E equals fraction numerator open parentheses 6 comma 63 cross times 10 to the power of negative 34 end exponent close parentheses open parentheses 3 cross times 10 to the power of 8 close parentheses over denominator 0 comma 07 cross times 10 to the power of negative 9 end exponent end fraction E equals 284 comma 14 cross times 10 to the power of negative 17 end exponent space straight J E equals 2 comma 84 cross times 10 to the power of negative 15 end exponent space straight J end style 

Maka, energi yang diberikan pada elektron
begin mathsize 14px style E subscript e equals E minus E apostrophe E subscript e equals open parentheses 2 comma 84 cross times 10 to the power of negative 15 end exponent close parentheses minus open parentheses 2 comma 79 cross times 10 to the power of negative 15 end exponent close parentheses E subscript e equals 0 comma 05 cross times 10 to the power of negative 15 end exponent space straight J end style 

Jadi, energi yang diberikan kepada elektron yang terpental sebesar begin mathsize 14px style bold 0 bold comma bold 05 bold cross times bold 10 to the power of bold minus bold 15 end exponent bold space bold J end style.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 14 Maret 2021

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