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Energi kinetik maksimum dari penyinaran logam kali...

Energi kinetik maksimum dari penyinaran logam kalium dengan ultraviolet (begin mathsize 14px style lambda equals 200 space n m end style) jika panjang gelombang ambang kalium 440 nm adalah...

  1. 3,0 eV

  2. 4,3 eV

  3. 3,9 eV

  4. 3,4 eV

  5. 9,9 eV

Jawaban:

Diketahui :

  • Konstanta Planck : begin mathsize 14px style h space equals space 6 comma 6 space x space 10 to the power of negative 34 end exponent space J s end style 
  • Kecepatan cahaya di ruang hampa : begin mathsize 14px style c space equals space 3 space x space 10 to the power of 8 space m divided by s end style 
  • Panjang gelombang : begin mathsize 14px style lambda equals 200 space n m equals 200 space x space 10 to the power of negative 9 end exponent space m end style 
  • Panjang gelombang ambang : begin mathsize 14px style lambda subscript 0 equals 440 space n m equals 440 space x space 10 to the power of negative 9 end exponent space m end style 

Energi kinetik maksimum :

begin mathsize 14px style E equals h c space open parentheses 1 over lambda minus 1 over lambda subscript 0 close parentheses E equals 6 comma 6 comma space x space 10 to the power of negative 34 end exponent left parenthesis 3 space x space 10 to the power of 8 right parenthesis space open parentheses fraction numerator 1 over denominator 200 space x space 10 to the power of negative 9 end exponent end fraction minus fraction numerator 1 over denominator 440 space x space 10 to the power of negative 9 end exponent end fraction close parentheses E equals 6 comma 6 comma space x space 10 to the power of negative 34 end exponent left parenthesis 3 space x space 10 to the power of 8 right parenthesis space open parentheses fraction numerator 2 comma 2 minus 1 over denominator 440 space x space 10 to the power of negative 9 end exponent end fraction close parentheses E equals left parenthesis 19 comma 8 space x space 10 to the power of negative 26 end exponent right parenthesis left parenthesis 0 comma 0027 space x space 10 to the power of 9 right parenthesis E equals 0 comma 054 space x space 10 to the power of negative 17 end exponent space J E equals fraction numerator 0 comma 054 space x space 10 to the power of negative 17 end exponent over denominator 1 comma 62 space x space 10 to the power of negative 19 end exponent end fraction equals 3 comma 4 space e V end style 

Jadi, jawaban yang tepat adalah D.

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