JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • question_answer
    How many elements would be in the second period of the Periodic Table if the spin quantum number \[{{m}_{s}}\] could have the value \[-\frac{1}{2},\,0,\,+\frac{1}{2}\]?

    A)  8                                

    B)  10              

    C)  12                              

    D)  18

    Correct Answer: C

    Solution :

     For second period n = 2 hence I           m         \[{{m}_{s}}\] 0          0          \[+\frac{1}{2},\,\,0,\,-\frac{1}{2}\] 1          - 1        \[+\frac{1}{2},\,0,\,-\frac{1}{2}\] 0          \[+\frac{1}{2},\,0,\,-\frac{1}{2}\] \[+1\]  \[+\frac{1}{2},\,0,\,-\frac{1}{2}\]  Hence, total number of elements = 12 (= total values of spin quantum number)

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