Question

C60 emerging from a source at a speed (v) has a de broglie wavelength of 11.0 a. The value of v (in ms-1) is closest to

Answer 1

Answer:

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Answer 2

Explanation:

It is given that,

The De Broglie wavelength, $\lambda=11\ A=11\times 10^{-10} m$

The De Broglie wavelength of an object is given by :

$\lambda=\dfrac{h}{mv}$

$v=\dfrac{h}{m\lambda}$

m is the mass of C - 60, m = 720.66 g/mol

In 1 mole C -60, $720\times 10^{-3}\ kg/mol$

Since, $1\ mole=6.022\times 10^{23}\ atoms$

So, 1 atom, $1\ atom=\dfrac{720\times 10^{-3}}{6.022\times 10^{23}}$

$v=\dfrac{6.63\times 10^{-34}\times 6.022\times 10^{23}}{720.66\times 10^{-3} \times 11\times 10^{-10}}$

v = 0.5 m/s

So, the speed of C - 60 is 0.5 m/s. Hence, this is the required solution.