JIPMER Jipmer Medical Solved Paper-1997

  • question_answer
    When an electron in hydrogen in hydrogen atom is excited, from its 4th to 5th stationary orbit. The change in angular momentum  of electron  is:  (Planck constant, \[h=6.6\,\times {{10}^{-34}}\,Js\])

    A)  \[2.09\text{ }Js\]                            

    B)  \[1.05\times {{10}^{-34}}Js\]     

    C)  \[3.32\times {{10}^{-34}}Js\]   

    D)         \[4.16\times {{10}^{-34}}Js\]

    Correct Answer: B

    Solution :

    Here: Number of initial orbit \[{{n}_{1}}=4\] Number of final orbit \[{{n}_{2}}=5\] Now change in angular momentum is given by \[\Delta L={{L}_{2}}-{{L}_{1}}=\frac{{{n}_{2}}h}{2\pi }-\frac{{{n}_{1}}h}{2\pi }\] \[=\frac{h}{2\pi }\left( {{n}_{2}}-{{n}_{1}} \right)\] \[=\frac{6.6\times {{10}^{-34}}}{2\times 3.14}(5-4)=1.05\times {{10}^{-34}}Js\]                


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