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Emiter dalam sebuah tabung fotolistrik memiliki pa...

Emiter dalam sebuah tabung fotolistrik memiliki panjang gelombang ambang 6.000 Å. Tentukan panjang gelombang cahaya yang datang pada tabung itu jika tegangan henti cahaya ini 2,5 V!undefined 

Jawaban:

begin mathsize 14px style lambda subscript 0 equals 6.000 space straight Å equals 6 cross times 10 to the power of negative 7 end exponent space straight m V subscript s equals 2 comma 5 space straight V end style 

Ditanyakan: panjang gelombang yang datang?
Jawab:

Energi kinetik maksimum
begin mathsize 14px style E k equals q. V subscript s E k equals open parentheses 1 comma 6 cross times 10 to the power of negative 19 end exponent close parentheses open parentheses 2 comma 5 close parentheses E k equals 4 cross times 10 to the power of negative 19 end exponent space straight J end style 

Panjang gelombang datang
begin mathsize 14px style E k equals h f minus h f subscript 0 E k equals h c over lambda minus h c over lambda subscript 0 4 cross times 10 to the power of negative 19 end exponent equals open parentheses 6 comma 63 cross times 10 to the power of negative 34 end exponent close parentheses fraction numerator open parentheses 3 cross times 10 to the power of 8 close parentheses over denominator lambda end fraction minus open parentheses 6 comma 63 cross times 10 to the power of negative 34 end exponent close parentheses fraction numerator open parentheses 3 cross times 10 to the power of 8 close parentheses over denominator open parentheses 6 cross times 10 to the power of negative 7 end exponent close parentheses end fraction 4 cross times 10 to the power of negative 19 end exponent equals fraction numerator open parentheses 19 comma 89 cross times 10 to the power of negative 26 end exponent close parentheses over denominator lambda end fraction minus 3 comma 315 cross times 10 to the power of negative 19 end exponent 7 comma 315 cross times 10 to the power of negative 19 end exponent equals fraction numerator open parentheses 19 comma 89 cross times 10 to the power of negative 26 end exponent close parentheses over denominator lambda end fraction lambda equals fraction numerator 19 comma 89 cross times 10 to the power of negative 26 end exponent over denominator 7 comma 315 cross times 10 to the power of negative 19 end exponent end fraction lambda equals 2 comma 719 space straight m lambda equals 2719 space straight Å end style 

Jadi, panjang gelombang cahaya yang datang sebesar 2719 Å.

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