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In H-atom if `r1` is the radius fo first Bohr orbit de-Broglie wavelength of an elecrton in ` 3^(rd)` orbit is :A. `3pi x`B. `6pi x`C. `(9x)/(2)`D. `x/2` |
Answer» Correct Answer - B `r_(1)=0.529 Å` `r_(3)=0.529xx(3)^(2)Å=9x` So, `lambda=(2pir)/(n)=(2pi(9x))/(3)=6 pi x` |
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