Question

Consider the hydrogen atom as described by the Bohr model. The nucleus of the hydrogen atom is a single proton. The electron rotates in a circular orbit about this nucleus. In the n = 5, orbit the electron is 1.32 10-9 m from the nucleus and it rotates with an angular speed of 3.30 1014 rad/s. Determine the electron's centripetal acceleration in m/s2.

Answer 1

Explanation:

It is given that the atom is hydrogen. And, its electron rotates in n = 5 orbit with angular speed $3.3 \times 10^{14} rad/s$.

Radius of circular path = $1.32 \times 10^{-9}$ m

Hence, first calculate the tangential speed as follows.

          Tangential speed = radius × angular speed

                                        = $4.356 \times 10^{5} m/s$

As formula to calculate time period is as follows.

                  T = $1.5211 \times 10^{-16} \times \frac{n^{3}}{z^{2}}$ sec

Also,     frequency $(\nu)$ = $\frac{1}{T}$

So, for n = 5 and z = 1 the value of frequency is as follows.

             frequency $(\nu)$ = $\frac{1}{1.5211 \times 10^{-16}} \times \frac{z^{2}}{n^{3}}$

                                   = $\frac{1}{1.5211 \times 10^{-16}} \times \frac{1}{(5)^{3}}$  

                                   = $5.259 \times 10^{13}$ Hz

As formula to calculate centripetal acceleration is as follows.

                  $a_{c} = \frac{v^{2}}{r}$

where,           v = linear speed

                      r = radius

                v = $2.165 \times 10^{6} \times \frac{z}{n}$

                   = $2.165 \times 10^{6} \times \frac{1}{5}$      

                   = $4.3 \times 10^{5} m/s$            

Hence, the centripetal acceleration will be as follows.

                 $a_{c} = \frac{v^{2}}{r}$

                            = $\frac{(4.3 \times 10^{5})^{2}}{1.32 \times 10^{-9}}$

                            = $1.4 \times 10^{20} m/s^{2}$

Thus, we can conclude that the electron's centripetal acceleration is $1.4 \times 10^{20} m/s^{2}$.