AIIMS AIIMS Solved Paper-2005

  • question_answer
    The most probable radius (in pm) for finding the electron in \[H{{e}^{+}}\] is:

    A) \[0.0\]                  

    B)        \[52.9\]

    C)        \[26.5\]               

    D)        \[105.8\]

    Correct Answer: C

    Solution :

    Bohrs radius \[(r)=\frac{59.2{{n}^{2}}}{Z}pm\] n = number of the shell Z = atomic number \[\text{r=}\frac{52.9\times {{1}^{2}}}{2}\text{ }pm=26.45\text{ }pm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner