• question_answer If ${{a}_{0}}$ be the radius of first Bohr's orbit of H-atom, the de-Broglie's wavelength of an electron revolving in the third Bohr's orbit will be A)  $6\pi {{a}_{0}}$                             B)  $4\pi {{a}_{0}}$ C)  $2\pi {{a}_{0}}$                             D)  $\pi {{a}_{0}}$

Idea This problem can be solved by using concept of Bohr's orbital parameter and de-Broglie equation, while solving the problem, students are advised to follow the following trend. Write the angular momentum equation$m\upsilon r=\frac{nh}{2\pi }$ Put the value of de-Broglie wavelength on this equation, then calculate value of $\lambda$ for 3rd orbit of H-atom.                 Use      ${{r}_{n}}={{n}^{2}}{{a}_{0}}$ Angular momentum,      $mv{{r}_{n}}=\frac{nh}{2\pi }$                ?(i) de-Broglie equation  $=p=\frac{h}{\lambda }=mv$        ...(ii) Placing the value of mv from Eqs. (ii) and (i) for 3rd orbit.                                 $\frac{h}{\lambda }{{r}_{3}}=\frac{3h}{2\pi }$                                 $\frac{\lambda }{h{{r}_{3}}}=\frac{2\pi }{3h}$                                 $\lambda =\frac{2\pi {{r}_{3}}}{3}$ $\because$     ${{r}_{3}}={{n}^{2}}{{a}_{0}}=9{{a}_{0}}$($\because$for 3rd orbit n = 3) So,          $\lambda =\frac{2\pi \cdot 9{{a}_{0}}}{3}=6\pi {{a}_{0}}$ TEST Edge Problems of same type involving the concept of Hesenberg uncertainty principle and for other nuclei such as $H{{e}^{+}},L{{i}^{2+}}$ are also asked in JEE Main, so students are advised to go through clear study of the above topics.
You will be redirected in 3 sec 