JEE Advance - Chemistry (2012 - Paper 1 Offline - No. 1)
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [$$\alpha_0$$ is Bohr radius]
$${h^2 \over {4{\pi ^2}m\alpha _0^2}}$$
$${h^2 \over {16{\pi ^2}m\alpha _0^2}}$$
$${h^2 \over {32{\pi ^2}m\alpha _0^2}}$$
$${h^2 \over {64{\pi ^2}m\alpha _0^2}}$$
Explanation
For Bohr orbit, angular momentum is
$$mvr_n = \frac{nh}{2\pi}$$
Velocity, $$v = \frac{nh}{2\pi mr_n} \quad \text{... (i)}$$
Kinetic energy, $$K.E. = \frac{1}{2}mv^2 \quad \text{... (ii)}$$
By putting the value of $v$ from (i) into (ii),
$$K.E. = \frac{1}{2}m \times \frac{n^2h^2}{4\pi^2m^2r_n^2} = \frac{n^2h^2}{8\pi^2mr_n^2}$$
For second Bohr orbit, $n = 2$
$$r_n = a_0 \times n^2 \quad (a_0 = \text{Bohr radius})$$
$$r_n = 4a_0$$
$$K.E. = \frac{(2)^2h^2}{8\pi^2m(4a_0)^2}$$
Thus, $$K.E. = \frac{h^2}{32\pi^2ma_0^2}$$
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