JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • 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)  \[\pi {{a}_{0}}\]

    Correct Answer: A

    Solution :

     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.

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