InterviewSolution
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InterviewSolution
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The radius of electron orbit and the speed of electron in the ground state of hydrogen atom is `5.30xx10^(-11)m` and `22xx10^(6)m s^(-1)` respectively, then the orbital peroid of this electron in second excited state will beA. `1.21xx10^(-14)s`B. `1.21xx10^(-12)s`C. `1.21xx10^(-10)s`D. `1.21xx10^(-15)s` |
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Answer» Correct Answer - D Here, `r_1=5.30xx10^(-11)m,v_1=2.2xx10^6ms^(-1)` In the second excited state, `r_n=n^2r_1,v_n=(v_1)/(n)` `therefore r_2=4r_1=4xx5.30xx10^(-11)m=2.12xx10^(-10)m` and `v_2=(v_1)/(2)=(2.2xx10^(6))/(2)ms^(-1)=1.1xx10^(6)ms^(-1)` |
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