Question

The electron in the hydrogen atom is moving with a speed
of 2.3 10° m/s in an orbit of radius 0:53Å. Calculate the
period of revolution of electron. (n = 3.142)​

Answer 1

v=2.3\times10^6m/sv=2.3×10

6

m/s

r=0.53 A^o=53\times10^{-8}mr=0.53A

o

=53×10

−8

m

T=?T=?

time taken by the electron to complete one revolution=T=\cfrac{circumference}{speed}=T=

speed

circumference

\therefore T=\cfrac{2\pi r}{v}=\cfrac{2\times3.142 \times53\times10^{-8}}{2.3\times10^6}=144.8\times10^{-14}sec∴T=

v

2πr

=

2.3×10

6

2×3.142×53×10

−8

=144.8×10

−14

sec