Physics - Modern Physics Question with Solution | TestHub

Acceleration of an electron in the $n^{\text{th}}$ orbit is given by $a=\frac{v^2}{r}$. Since $v\propto \frac{1}{n}$ and $r\propto n^2$, we have $a\propto \frac{1}{n^4}$. Thus, acceleration in $n=2$ is less than in $n=1$.

Angular momentum is $L=\frac{nh}{2\pi }\propto n$. Thus, angular momentum in $n=2$ is more than in $n=1$.

Kinetic energy is $K=\frac{13.6}{n^2}\text{ eV}\propto \frac{1}{n^2}$. Thus, kinetic energy in $n=2$ is less than in $n=1$.

Potential energy is $U=-\frac{27.2}{n^2}\text{ eV}$. For $n=1$, $U_1=-27.2\text{ eV}$. For $n=2$, $U_2=-6.8\text{ eV}$.

Since $-6.8\text{ eV}>-27.2\text{ eV}$, the potential energy in $n=2$ is greater than in $n=1$.

Therefore, the statement that potential energy in $n=2$ orbit is less than that in $n=1$ orbit is false.

Answer: Potential energy of electron in $\mathrm{n}=2$ orbit is less than that in $\mathrm{n}=1$ orbit.