The uncertainity in velocity of an electron present in the nucleus of diameter 10^-10 m hypothetically should be approximately
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Hello,
here you can use the formula,
∆x ×∆p=h/4π
(Heisenberg's uncertainty principle)
∆v=h/4πm∆x
∆v=6.63×10^-34÷(10^-10×9.1×10^-31×4×3.14
∆v=0.058×10^-13
∆v=5.8×10^-15m/s
Hope that helps.
keep learning.
here you can use the formula,
∆x ×∆p=h/4π
(Heisenberg's uncertainty principle)
∆v=h/4πm∆x
∆v=6.63×10^-34÷(10^-10×9.1×10^-31×4×3.14
∆v=0.058×10^-13
∆v=5.8×10^-15m/s
Hope that helps.
keep learning.
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