Sebuah elektron dengan massa 9,11×10−31kg dan muatan listrik −1,6×10−19C, lepas dari katode menuju ke anode yang jaraknya 2 cm. Jika kecepatan awal elektron 0 m/s dan beda potensial antara anode dan katode 200 V, maka elektron akan sampai dianode dengan kecepatan ....

Pertanyaan

Sebuah elektron dengan massa $9,11\times10^{-31}\;\mathrm{kg}$ dan muatan listrik $-1,6\times10^{-19}\;\mathrm C$ , lepas dari katode menuju ke anode yang jaraknya 2 cm. Jika kecepatan awal elektron 0 m/s dan beda potensial antara anode dan katode 200 V, maka elektron akan sampai dianode dengan kecepatan .... $\;$

$2 comma 3 cross times 10 to the power of 5 space straight m divided by straight s$
$8 comma 4 cross times 10 to the power of 6 space straight m divided by straight s$
$2 comma 3 cross times 10 to the power of 7 space straight m divided by straight s$
$3 cross times 10 to the power of 7 space straight m divided by straight s$
$2 comma 4 cross times 10 to the power of 8 space straight m divided by straight s$

Pembahasan Soal:

Diketahui :

$table attributes columnalign right center left columnspacing 0px end attributes row m equals cell 9 comma 11 cross times 10 to the power of negative 31 end exponent space kg end cell row q equals cell negative 1 comma 6 cross times 10 to the power of negative 19 end exponent space straight C end cell row d equals cell 2 space cm end cell row cell v subscript k a t o d a end subscript end cell equals cell 0 space straight m divided by straight s end cell row cell V subscript a n o d a minus k a t o d a end subscript end cell equals cell 200 space straight V end cell end table$

Ditanya :

$v subscript a n o d a end subscript equals.... ?$

Jawab :

Kita gunakan konsep energi mekanik. Energi mekanik merupakan penjumlahan antara energi potensial dan energi kinetik.

Sehingga berlaku persamaan :

$table attributes columnalign right center left columnspacing 0px end attributes row cell E M subscript k a t o d a end subscript end cell equals cell E M subscript a n o d a end subscript end cell row cell E P subscript k a t o d a end subscript plus E K subscript k a t o d a end subscript end cell equals cell E P subscript a n o d a end subscript plus E K subscript a n o d a end subscript end cell row cell q V subscript k a t o d a end subscript plus 1 half m v squared subscript k a t o d a end subscript end cell equals cell q V subscript a n o d a end subscript plus 1 half m v squared subscript a n o d a end subscript end cell row cell q V subscript k a t o d a end subscript minus q V subscript a n o d a end subscript end cell equals cell 1 half m v squared subscript a n o d a end subscript minus 1 half m v squared subscript k a t o d a end subscript end cell row cell q open parentheses V subscript k a t o d a end subscript minus V subscript a n o d a end subscript close parentheses end cell equals cell 1 half m v squared subscript a n o d a end subscript minus 1 half m v squared subscript k a t o d a end subscript end cell row cell q V subscript k a t o d a minus a n o d a end subscript end cell equals cell 1 half m v squared subscript a n o d a end subscript minus 1 half m open parentheses 0 close parentheses end cell row cell q V subscript k a t o d a minus a n o d a end subscript end cell equals cell 1 half m v squared subscript a n o d a end subscript end cell row cell open parentheses negative 1 comma 6 cross times 10 to the power of negative 19 end exponent close parentheses open parentheses negative 200 close parentheses end cell equals cell open parentheses 1 half close parentheses open parentheses 9 comma 11 cross times 10 to the power of negative 31 end exponent close parentheses v squared subscript a n o d a end subscript end cell row cell v squared subscript a n o d a end subscript end cell equals cell fraction numerator open parentheses 1 comma 6 cross times 10 to the power of negative 19 end exponent close parentheses open parentheses 200 close parentheses over denominator open parentheses 1 half close parentheses open parentheses 9 comma 11 cross times 10 to the power of negative 31 end exponent close parentheses end fraction end cell row cell v squared subscript a n o d a end subscript end cell equals cell fraction numerator open parentheses 1 comma 6 cross times 10 to the power of negative 17 end exponent close parentheses open parentheses 4 close parentheses over denominator open parentheses 9 comma 11 cross times 10 to the power of negative 31 end exponent close parentheses end fraction end cell row cell v squared subscript a n o d a end subscript end cell equals cell fraction numerator open parentheses 1 comma 6 close parentheses open parentheses 4 close parentheses open parentheses 10 to the power of 14 close parentheses over denominator open parentheses 9 comma 11 close parentheses end fraction end cell row cell v subscript a n o d a end subscript end cell equals cell 0 comma 84 cross times 10 to the power of 7 space straight m divided by straight s end cell row cell straight v subscript anoda end cell equals cell 8 comma 4 cross times 10 to the power of 6 space straight m divided by straight s end cell end table$

Dengan demikian, kecepatan di katoda adalah $8 comma 4 cross times 10 to the power of 6 space straight m divided by straight s$ .

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

J. Khairina

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 07 Oktober 2021