InterviewSolution
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InterviewSolution
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The de-Broglie wavelength of an electron in the first Bohr orbit isA. equal to one- fourth the circumference of the first orbitB. equal to half the circumference of first orbitC. equal to twice the circumference of first orbit.D. equal to the circumference of the first orbit. |
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Answer» Correct Answer - D Angular momentum `=(nh)/(2pi)` `rArr` moment of momentum `=pxxr_n` `rArrpxxr_n=(nh)/(2pi)rArrr(h)/(lambda)r_nrArrlambda =(2pir_n)/(n)` For `1^st` orbit, `n=1,lambda =2pir_1` `rArr =` circumference of `1^st` orbit. |
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