A) \[{{\lambda }_{B}}={{\lambda }_{G/3}}\]
B) \[{{\lambda }_{B}}={{\lambda }_{G/2}}\]
C) \[{{\lambda }_{B}}=2{{\lambda }_{G}}\]
D) \[{{\lambda }_{B}}=3{{\lambda }_{G}}\]
Correct Answer: D
Solution :
de-Broglie wavelength \[\lambda =\frac{h}{mv}\] \[mvr=\frac{nh}{2\pi }\] \[\frac{h}{mv}=\frac{2\pi r}{n}\] \[\lambda =\frac{2\pi r}{n}\] \[r={{a}_{0}}\frac{{{n}^{2}}}{Z}\] \[\lambda =\frac{2\pi {{a}_{0}}n}{Z}\] As the atom is hydrogen Z = 1 \[{{\lambda }_{B}}=\frac{2\pi {{a}_{0}}3}{Z}\] \[{{\lambda }_{G}}=\frac{2\pi a0}{Z}\] \[{{\lambda }_{B}}=3{{\lambda }_{G}}\]You need to login to perform this action.
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