Roboguru

Panjang gelombang ambang suatu logam perak . Jika logam tersebut disinari dengan sinar ultraviolet dengan panjang gelombang . Hitung: fungsi kerja energi kinetik maksimum elektron yang lepas

Pertanyaan

Panjang gelombang ambang suatu logam perak begin mathsize 14px style 3.750 space straight A with degree on top end style. Jika logam tersebut disinari dengan sinar ultraviolet dengan panjang gelombang begin mathsize 14px style 2.500 space straight A with degree on top end style. Hitung:

  1. fungsi kerja
  2. energi kinetik maksimum elektron yang lepas

Pembahasan Soal:

Diketahui

begin mathsize 14px style lambda with degree on top equals 3750 space straight A with degree on top equals 3 comma 75 cross times 10 to the power of negative 7 space end exponent straight m lambda equals 2500 space straight A with degree on top equals 2 comma 5 cross times 10 to the power of negative 7 space end exponent straight m end style

Ditanyakan begin mathsize 14px style ϕ space dan space E K end style

Penyelesaian

untuk menentukan fungsi kerja gunakan persamaan berikut

begin mathsize 14px style ϕ equals fraction numerator h c over denominator begin display style lambda with degree on top end style end fraction ϕ equals fraction numerator 6 comma 63 cross times 10 to the power of negative 34 end exponent times 3 cross times 10 to the power of 8 over denominator begin display style 3.75 cross times 10 to the power of negative 7 end exponent end style end fraction ϕ equals 5 comma 3 cross times 10 to the power of negative 19 end exponent space J ϕ equals fraction numerator 5 comma 3 cross times 10 to the power of negative 19 end exponent over denominator begin display style 1 comma 6 cross times 10 to the power of negative 19 end exponent end style end fraction ϕ equals 3 comma 31 space eV end style

adapun untuk menentukan energi kinetik maksimum gunakan persamaan Efek Fotolistrik

begin mathsize 14px style E k subscript m a x end subscript equals fraction numerator h c over denominator lambda end fraction minus ϕ E k subscript m a x end subscript equals fraction numerator 6 comma 63 cross times 10 to the power of negative 34 end exponent times 3 cross times 10 to the power of 8 over denominator 2 comma 5 cross times 10 to the power of negative 7 end exponent end fraction minus 3 comma 31 space eV E k subscript m a x end subscript equals 7 comma 96 cross times 10 to the power of negative 19 end exponent space straight J minus 3 comma 31 space eV E k subscript m a x end subscript equals fraction numerator 7 comma 96 cross times 10 to the power of negative 19 end exponent space over denominator 1 comma 6 cross times 10 to the power of negative 19 end exponent end fraction minus 3 comma 31 space eV E k subscript m a x end subscript equals 4 comma 98 space eV minus 3 comma 31 space eV E k subscript m a x end subscript equals 1 comma 67 space eV end style

Dengan demikian besar fungsi kerja adalah 3,31 eV sedangkan besar energi kinetik maksimum adalah 1,67 eV

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 15 Maret 2021

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