Numericals for hinesberg uncertainty principal

Numericals
3 calculate the product of uncertainty in displacement and velocity for an electron with mass 9.1×10^-31kg.
sol.
mass of electron =9.1×10^-31kg

h= 6.6×10^-34kgm^3/s

now

we know

∆x.∆p= h/4π

∆x.m∆V=h/4π

∆x.∆V =h/(4πm)

putting all value there

∆x.∆V= 5.77×10^-5 m^2/s ans

4. On the basis of uncertainty principal so that electron cannot exist in the nucleus.
sol.
radius of nucleus = 10^-14 m

h= 6.6×10^-34kgm^3/s

mass of electron =9.1×10^-31kg

we know
∆x.∆p= h/4π

∆x.m∆V=h/4π

∆V =h/(4πm∆x)

putting all value

we get
∆V= 5.77×10^9m/s

since this value 5.77×10^9m/s is greater than
velocity 3×10^8m/s which is not possible. hence electron cannot be found in the nucleus of atom.

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