27. the kinetic energy of an electron in the second bohr orbit of a hydrogen atom is a₀ is bohr radius…
27. the kinetic energy of an electron in the second bohr orbit of a hydrogen atom is a₀ is bohr radius : 2012 (a) $\frac{h^{2}}{4pi^{2}ma_{0}^{2}}$ (b) $\frac{h^{2}}{16pi^{2}ma_{0}^{2}}$ (c) $\frac{h^{2}}{32pi^{2}ma_{0}^{2}}$ (d) $\frac{h^{2}}{64pi^{2}ma_{0}^{2}}$
Answer
Explanation:
Step1: Recall Bohr's formula for velocity
The velocity of an electron in the $n^{th}$ orbit of a hydrogen - like atom is given by $v = \frac{Z e^{2}}{2 \epsilon_{0} n h}$. For hydrogen ($Z = 1$), in the second orbit ($n = 2$), $v=\frac{e^{2}}{4\epsilon_{0}h}$. Also, from the quantization of angular momentum $mvr=\frac{nh}{2\pi}$, and for hydrogen - like atoms, the radius of the $n^{th}$ orbit is $r = n^{2}a_{0}\frac{1}{Z}$, for hydrogen $Z = 1$ and $n = 2$, so $r = 4a_{0}$. From $mvr=\frac{nh}{2\pi}$, we can also express $v=\frac{nh}{2\pi mr}$. Substituting $n = 2$ and $r = 4a_{0}$, we get $v=\frac{h}{4\pi ma_{0}}$.
Step2: Calculate kinetic energy
The kinetic energy of an electron is $K.E.=\frac{1}{2}mv^{2}$. Substitute $v=\frac{h}{4\pi ma_{0}}$ into the kinetic - energy formula: [ \begin{align*} K.E.&=\frac{1}{2}m\left(\frac{h}{4\pi ma_{0}}\right)^{2}\ &=\frac{h^{2}}{32\pi^{2}ma_{0}^{2}} \end{align*} ]
Answer:
(c) $\frac{h^{2}}{32\pi^{2}ma_{0}^{2}}$