Question

What is the de-Brogile wavelength of fullerene $\left(C_{60}\right)$, if it is moving with $4\times {10}^5\text{m/s}$?
$(N_A=6\times {10}^{23})$

• A. $1.38\times {10}^{-15}\text{m}$
• B. $1.38\times {10}^{-13}\text{m}$
• C. $8.26\times {10}^{-13}\text{m}$
• D. $8.26\times {10}^{10}\text{m}$

Answer 1

The correct option is A $1.38\times {10}^{-15}\text{m}$

$\text{Mass of}1C_{60}\text{molecule}$
$=\frac{60\times 12}{6\times {10}^{23}\times {10}^3}=1.2\times {10}^{-24}\text{kg}$
Given, velocity $V=4\times {10}^5\text{m/s}$
So, de Broglie wavelength$\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-34}}{1.2\times {10}^{-24}\times 4\times {10}^5}$
$=1.38\times {10}^{-15}\text{m}$