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An electron is revolving round the nucleus of `He^(+)` ion with speed `2.2 xx 10^(6) m//s`. The potential energy of the electron is (if atomic number of `He = 2, 1 eV = 16 xx 10^(-19) J`. Mass of electron `= 9 xx 10^(-31) kg` )A. `- 13.61 eV`B. `- 6.8 eV`C. `- 27.22 eV`D. zero |
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Answer» Correct Answer - C `KE = (1)/(2)m xx v^(2) = (1)/(2) xx 9 xx 10^(-31) xx (2.2 xx 10^(6))^(2)` `KE = (1)/(2) xx (9 xx 10^(-31) xx (2.2 xx 10^(6))^(2))/(1.6 xx 10^(-19))` `KE = 13.61 eV` `PE = -2 xx KE = -2 xx 13.61 = -27.22 eV` |
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