C_{60} emerging from a source at a speed (v) | vedclass

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(A) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.First,calculate the mass of one $C_{60}$ molecule in $kg$:Mass of $C_{6

Vedclass · Accepted Answer

$0.5$

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(A) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.First,calculate the mass of one $C_{60}$ molecule in $kg$:Mass of $C_{60} = 60 	imes 12 \ 	ext{amu} = 720 \ 	ext{amu}$.Since $1 \ 	ext{amu} = 1.66 	imes 10^{-27} \ kg$,the mass $m = 720 	imes 1.66 	imes 10^{-27} \ kg \approx 1.195 	imes 10^{-24} \ kg$.Given $\lambda = 11.0 \ \mathring{A} = 11.0 	imes 10^{-10} \ m$ and $h = 6.626 	imes 10^{-34} \ J \ s$.Rearranging for $v$: $v = \frac{h}{m \lambda} = \frac{6.626 	imes 10^{-34}}{1.195 	imes 10^{-24} 	imes 11.0 	imes 10^{-10}}$.$v \approx \frac{6.626 	imes 10^{-34}}{1.3145 	imes 10^{-33}} \approx 0.504 \ m \ s^{-1}$.Comparing this with the given options,the value is closest to $0.5 \ m \ s^{-1}$,which corresponds to option $(A)$.

$C_{60}$ emerging from a source at a speed $(v)$ has a de Broglie wavelength of $11.0 \ \mathring{A}$. The value of $v$ (in $m \ s^{-1}$) is closest to
[Planck's constant $h = 6.626 \times 10^{-34} \ J \ s$]

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$C_{60}$ emerging from a source at a speed $(v)$ has a de Broglie wavelength of $11.0 \ \mathring{A}$. The value of $v$ (in $m \ s^{-1}$) is closest to[Planck's constant $h = 6.626 \times 10^{-34} \ J \ s$]