Question

Ratio of time period of revolution of electron in 1st orbit of He+ ion and 2nd orbit of Be3+ ion, respectively

Answer 1

Answer: Ratio is 1/2

Explanation: By taking the Time period of revolution of electron as the circumference of the orbit divided by the speed of the moving electron.  Thus $T = 2\pi r/v$ where T is Time period, r is Radius of orbit, v is Velocity of electron.

Now we know that r is directly proportional to $n^2/Z$ according to the equations of Bohr's atom and v is directly proportional to Z/n, where n is the orbit number and Z is the atomic number of the atoms.  

Now taking the ratio of $T_1/T_2$ we get $n^3_1/n^3_2 * Z^2_2/Z^2_1$.

We know that the value of  $n_1 = 1$ for He atom and $n_2 = 2$ for Be atom and $Z_1 = 2$  for He atom and $Z_2 = 4$ for Be atom.

By substituting the given values in the ratio we get $T_1/T_2 = 1/2$