Question

The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6\times {10}^{-16}s.$ The frequency of revolution of the electron in its first excited state (in $s^{-1}$) is :

• A. $6.2\times {10}^{15}$
• B. $1.6\times {10}^{14}$
• C. $7.8\times {10}^{14}$
• D. $5.6\times {10}^{12}$

Answer 1

The correct option is C $7.8\times {10}^{14}$

Time period is proportional to $\frac{n^3}{Z^2}$.
Let $T_1$ be the time period in ground state and $T_2$ be the time period in it’s first excited state.
$$T_1=\frac{n^3}{2^2}$$
(Where, n = excitation level and 2 is atomic no.)
$$\frac{T_1}{T_2}={\left(\frac{n_1}{n_2}\right)}^3$$
Given, $$T_1=1.6\times {10}^{-16}s$$
So, $$\frac{1.6\times {10}^{-16}}{T2}={\left(\frac{1}{2}\right)}^3$$
$$T_2=12.8\times {10}^{-16}s$$
Frequency is given by $f=\frac{1}{T}$
$$f=\frac{1}{12.8}\times {10}^{16}Hz$$
$$f=7.8128\times {10}^{14}Hz$$