IIT-JEE 2012 · Paper 1 · Chemistry Q28

ChemistrySingle correct+3 / -1Question 28 of 60

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ $a_0$ is Bohr radius]

\dfrac{h^2}{4\pi^2ma_0^2}

\dfrac{h^2}{16\pi^2ma_0^2}

\dfrac{h^2}{32\pi^2ma_0^2}

\dfrac{h^2}{64\pi^2ma_0^2}

Answer and worked solution

Official answer: C

Bohr model. $mvr=\dfrac{nh}{2\pi}$ and $r_n=n^2a_0$ . For $n=2$ , $r=4a_0$ :
$v=\frac{2h}{2\pi m(4a_0)}=\frac{h}{4\pi ma_0}.$

Kinetic energy.
$K=\frac12mv^2=\frac12m\cdot\frac{h^2}{16\pi^2m^2a_0^2}=\frac{h^2}{32\pi^2ma_0^2}.$

(C)

Topics:Bohr model

IIT-JEE 2012 · Paper 1, Chemistry question 28