Question

Suppose a single electron orbits about a nucleus containing two protons (+2e), as would be the case for a helium atom from which one of the naturally occurring electrons is removed. The radius of the orbit is 2.99 × 10-11 m. Determine the magnitude of the electron's centripetal acceleration.

Answer 1

Answer:

$a=5.66*10^{23} \frac{m}{s^2}$

Explanation:

In this case we will use the Bohr Atomic model.

We have that: $F=m*a$

We can calculate the centripetal force using the coulomb formula that states:

$F=k*\frac{q*q'}{r^2}$

Where K=$9*10^9 \frac{Nm^2}{C}$

and r is the distance.

Now we can say:

$m*a=k*\frac{q*q'}{r^2}$

The mass of the electron is = $9.1*10^{-31}$ Kg

The charge magnitud of an electron and proton are= $1.6*10^{-19}C$

Substituting what we have:

$[tex]a=\frac{9*10^{9}*(1.6*10^{-19} )*(2(1.6*10^{-19} ))}{9.1*10^{-31}*(2.99*10^{-11})^2 }$[/tex]

so:

$a=5.66*10^{23} \frac{m}{s^2}$