In the ground state ofhydrogen, the uncertainty in the position of the electron
ID: 1758045 • Letter: I
Question
In the ground state ofhydrogen, the uncertainty in the position of the electron isroughly 0.15nm. The mass of ane electron is 9.10939 × 1031 kg and thePlanck’s constant is 6.62607 × 1034 J· s. If thespeed of the electron is on the order of the uncertainty in thespeed, how fast is the electron moving? Answer in units ofm/s. I tried this method, but I think I'mdoing something wrong. Please check! Could the steps please beexplained, thanks. ( h/ 2 ) / [ m dx ]= 771,784 m / s In the ground state ofhydrogen, the uncertainty in the position of the electron isroughly 0.15nm. The mass of ane electron is 9.10939 × 1031 kg and thePlanck’s constant is 6.62607 × 1034 J· s. If thespeed of the electron is on the order of the uncertainty in thespeed, how fast is the electron moving? Answer in units ofm/s. I tried this method, but I think I'mdoing something wrong. Please check! Could the steps please beexplained, thanks. ( h/ 2 ) / [ m dx ]= 771,784 m / s I tried this method, but I think I'mdoing something wrong. Please check! Could the steps please beexplained, thanks. ( h/ 2 ) / [ m dx ]= 771,784 m / sExplanation / Answer
From the uncertainity in position and momentum we have : x P ˜ h /4 x (mv) ˜ h /4 (since momentum P = mass * velocity = m v) x m (v)˜ h /4 or v = h /4 *(x * m ) Given : x = 0.15 nm = 0.15 x 10-9 m h = 6.626 x 10-34 J-s m = 9.11 x 10-31 kg v = (6.626 x 10-34 J-s ) /4*3.14*( 0.15 x 10-9 m * 9.11 x 10-31 kg ) = 0.386057 x 106 m /s = 386,057 m/s Hope this helps u!Related Questions
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