Question

If the kinetic energy of a sub-atomic particle is ${3.313\times 10}^{-27}J$, then the frequency of particle wave will be:

$(Given:h={6.626\times 10}^{-34}Jsec)$

• A. ${10}^5{sec}^{-1}$
• B. ${10}^7{sec}^{-1}$
• C. ${10}^8{sec}^{-1}$
• D. ${10}^{10}{sec}^{-1}$

Answer 1

The correct option is B ${10}^7{sec}^{-1}$

$p=\sqrt{2\times k\times m}$

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

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

$\nu =\frac{C}{\lambda }$

$\frac{1}{\lambda }=\frac{\nu }{c}$

$p=\frac{h\nu }{c}$

$\frac{h\nu }{c}=\sqrt{2\times k\times m}$

$\nu =\frac{c\times \sqrt{2\times k\times m}}{h}$

$=\frac{3\times {10}^8\times \sqrt{2\times 3.313\times {10}^{-27}\times 9.1\times {10}^{-31}}}{6.626\times {10}^{-34}}$

KE = $\frac{1}{2}m{v}^2$

$2KE=mv.v$

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

$2K.E=\frac{h}{\lambda }\cdot v$

$v$ = velocity

$v=\lambda \nu$

$c=\lambda \nu$

$\therefore 2KE=\frac{h}{\lambda }\cdot \lambda \nu$

$2KE=h\nu$

$KE=\frac{h\nu }{2}$

$\nu =\frac{2KE}{h}$

$=\frac{2\times 3.313\times {10}^{-27}}{6.626\times {10}^{-34}}$

$={10}^7{\sec }^{-1}$