The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be

<p>According to Bohr&#39;s model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck&#39;s constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck&#39;s constant). This can be expressed as:</p> <p>$L = n \frac{h}{2\pi}$</p> <p>where $n$ is the principal quantum number or the orbit number.</p> <p>So, for the first orbit ($n=1$), the angular momentum $L_1$ is:</p> <p>$L_1 = 1 \times \frac{h}{2\pi} = \frac{h}{2\pi}$</p> <p>And for the second orbit ($n=2$), the angular momentum $L_2$ is:</p> <p>$L_2 = 2 \times \frac{h}{2\pi} = \frac{h}{\pi}$</p> <p>The change in angular momentum when moving from the first to the second orbit is the difference between $L_2$ and $L_1$:</p> <p>$\Delta L = L_2 - L_1 = \frac{h}{\pi} - \frac{h}{2\pi} = \frac{h}{2\pi}$</p> <p>Since $\frac{h}{2\pi}$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.</p> <p>Therefore, the correct answer is $L$.</p>