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प्रश्न
Calculate the de-Broglie wavelength of the electron orbitting in the n = 2 state of hydrogen atom.
उत्तर
As we know,
`mvr_2=(nh)/(2pi)`
The de-Broglie wavelength is given as,
`lambda=h/(mv)=h/(nh)xx(2pir_2)=pir_2`
As r2=0.529×n2×10−10 m
Therefore, λ=π×0.529×10−10×n2=3.14×0.529×10−10×22≃6.64×10−10 m
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