Question

$\\ \huge{\underline{\sf Chemistry\:Question:-}}$

If K.E of electron is 2.5×10^-24J.Then calculate the de-brogile wavelength.​

Answer 1

Answer:-

• Kinetic Energy=$\rm \dfrac{1}{2}mv^2$

Here

• KE=$\rm 2.5\times 10^{-24}J$
• Mass of electron=m=$\rm 9.1\times 10^{-31}kg$
• Wavelength=$\lambda$=?

We know

$\boxed{\nu =\left(\dfrac{2KE}{m}\right)^{\frac{1}{2}}}$

• Substitute the values.

$\\ \rm\longmapsto \nu=\left(\dfrac{2\times 2.5\times 10^{-24}J}{9.1\times 10^{-31}kg}\right)^{\frac{1}{2}}$

$\\ \rm\longmapsto \nu=2.34\times 10^3m/s$

Now

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

$\\ \rm\longmapsto \lambda=\dfrac{6.626\times 10^{-34}Js}{(9.1\times 10^{-31}kg)(2.34\times 10^3ms^{-1}}$

$\\ \rm\longmapsto \lambda=311.1\times 10^{-9}m$

$\\ \bf\longmapsto \lambda=311.1nm$