Question

$C_{60}$ 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, $h=6.626\times {10}^{-34}$ Js]

• A. $0.5$
• B. $2.5$
• C. $5.0$
• D. $30$

Answer 1

Answer: (A)

$0.5$

1 amu $=1.66054\times {10}^{-27}$ kg

Mass of 1 C atom $=12$ amu.

Mass of $C_{60}=60\times 12\times 1.66054\times {10}^{-27}kg=1.1956\times {10}^{-24}kg$

The de Broglie wavelength is given by the expression:

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

$11.0\mathring{A}\times {10}^{-10}m/\mathring{A}=\frac{6.626\times {10}^{-34}Js}{1.1956\times {10}^{-24}kg\times V}$

$V=0.5m/s$

Thus, the velocity is 0.5 m/s.

Hence, option A is correct.