The time period of revolution of an electron | vedclass

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(A) According to Bohr's theory,the radius of the $n^{	ext{th}}$ orbit is $r_n = r_0 \frac{n^2}{Z}$ and the velocity of the electron is $v_n = v_0 \fr

Vedclass · Accepted Answer

$T_0 = 1.5 	imes 10^{-16} \, s, a = 3$

Vedclass · Answer

(A) According to Bohr's theory,the radius of the $n^{	ext{th}}$ orbit is $r_n = r_0 \frac{n^2}{Z}$ and the velocity of the electron is $v_n = v_0 \frac{Z}{n}$.The time period $T$ is given by $T = \frac{2 \pi r_n}{v_n}$.Substituting the expressions for $r_n$ and $v_n$,we get $T = \frac{2 \pi (r_0 n^2 / Z)}{v_0 Z / n} = \frac{2 \pi r_0}{v_0} \cdot \frac{n^3}{Z^2}$.Comparing this with the given formula $T = \frac{T_0 n^a}{Z^b}$,we find $a = 3$ and $b = 2$. However,in standard physics problems of this type,the formula is often simplified to $T = T_0 \frac{n^3}{Z^2}$. Given the options provided,the closest match for the constant $T_0$ (which is $\frac{2 \pi r_0}{v_0}$) is approximately $1.51 	imes 10^{-16} \, s$ with $a = 3$ and $b = 2$. Since $b=2$ is not explicitly in the options,we select the option that correctly identifies the constant $T_0$ and the exponent $a=3$.

The time period of revolution of an electron in the $n^{\text{th}}$ orbit in a hydrogen-like atom is given by $T = \frac{T_0 n^a}{Z^b}$,where $Z$ is the atomic number. Identify the correct values for $T_0$,$a$,and $b$.

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