CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is...... (radius of the first orbit of hydrogen atom =0.53\[{{A}^{o}}\]).

    A)  \[1.67\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B) \[3.33\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) \[1.06\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D) \[0.53\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: B

    Solution :

    According to Bohrs quantisation of angular momentum \[mvr=\frac{nh}{2\pi }\]or\[\frac{h}{mv}=\frac{2\pi r}{n}\] ?(i) de- Broglie wavelength\[\lambda =\frac{h}{mv}\] ?(ii) From Eqs. (i) and (ii), we get Wavelength \[\lambda =\frac{2\pi r}{n}\] \[=\frac{2\times \pi \times 0.53{\AA}}{1}\] \[=3.33{\AA}\]


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