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Hitung jari-jari orbit, kecepatan, momentum sudut; dan percepatan sentripetal elektron dalam orbit n = 2 atom hidrogen?

Pertanyaan

Hitung jari-jari orbit, kecepatan, momentum sudut; dan percepatan sentripetal elektron dalam orbit n = 2 atom hidrogen?space   

Pembahasan Soal:

open vertical bar E subscript p close vertical bar equals open vertical bar fraction numerator negative 27 comma 2 over denominator n squared end fraction e V close vertical bar equals open vertical bar fraction numerator k q subscript e squared over denominator r end fraction close vertical bar r equals fraction numerator n squared over denominator left parenthesis 27 comma 2 e V right parenthesis k q subscript e squared end fraction equals fraction numerator left parenthesis 2 right parenthesis squared over denominator left parenthesis 27 comma 2 cross times 1 comma 6 cross times 10 to the power of negative 19 end exponent right parenthesis left parenthesis 9 cross times 10 to the power of 9 right parenthesis left parenthesis 1 comma 6 cross times 10 to the power of negative 19 end exponent right parenthesis end fraction r equals 2 comma 12 cross times 10 to the power of negative 10 end exponent space m  v equals fraction numerator 2 pi k q subscript e squared over denominator n h end fraction equals fraction numerator 2 pi left parenthesis 9 cross times 10 to the power of 9 right parenthesis left parenthesis 1 comma 6 cross times 10 to the power of negative 19 end exponent right parenthesis squared over denominator 2 left parenthesis 6 comma 6 cross times 10 to the power of negative 34 end exponent right parenthesis end fraction v equals 1 comma 09 cross times 10 to the power of 6 space m divided by s  L equals fraction numerator n h over denominator 2 pi end fraction equals fraction numerator 2 left parenthesis 6 comma 6 cross times 10 to the power of negative 34 end exponent right parenthesis over denominator 2 pi end fraction L equals 2 comma 11 cross times 10 to the power of negative 34 end exponent space k g m squared divided by s  a subscript s equals v squared over r equals fraction numerator left parenthesis 1 comma 09 cross times 10 to the power of 6 right parenthesis squared over denominator 2 comma 12 cross times 10 to the power of negative 10 end exponent end fraction a subscript s equals 5 comma 63 cross times 10 to the power of 21 space m divided by s squared       

Sehingga, jawaban yang tepat adalah 2,12 X 10-10 m; 1,09 X 106 m/ s; 2,11 X 10-34 kg m2/s; 5,63 X 1021 m/s2

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 13 Maret 2021

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