Question

if the ke of electron is 2.5×10^-24 J, then calculate its de-broglie wavelength

Answer 1

$\text{\underline{We\:know\:that:}}$

$\sf{Kinetic \: energy = \frac{1}{2}\: {mv}^{2}}$

$\sf\boxed{v = ( \frac{2 \: KE}{m} ) ^{ \frac{1}{2} }}$


$\text{\underline{Here,}}$

$\sf{Mass \: of \: electron = 9.1 \times {10}^{ - 31} kg}$

$\sf{KE = 2.5 \times {10}^{ - 24} J}$

$\sf{m = 9.1 \times {10}^{ - 31} kg}$


$\text{\underline{By\:substituting\:the\:values,}}$

$\text{\underline{We\:get,}}$

$\sf{v = ( \frac{2 \times 2.5 \times {10}^{ - 24} J}{9.1 \times {10}^{ - 31}kg } ) ^{ \frac{1}{2} }}$

$\sf\boxed{v = 2.34 \times {10}^{3} {ms}^{ - 1}}$


$\text{\underline{NOTE:}}$

$\boxed{1 \: J = kg \: {m}^{2} \: {s}^{ - 2}}$


$\text{\underline{We\:know\:that:}}$

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


$\text{\underline{Here,}}$

$\sf{h = 6.626 \times {10}^{ - 34} Js}$

$\sf{m = 9.1 \times {10}^{ - 31} kg}$

$\sf{v = 2.34 \times {10}^{3} {ms}^{ - 1}}$


$\text{\underline{By\:substituting\:the\:values,}}$

$\text{\underline{We\:get,}}$

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

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

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

$\boxed{\lambda = 311.1 \: nm}$


$\text{\underline{NOTE:}}$

$\boxed{1 \: nm = {10}^{ - 9} m}$

Answer 2

Explanation:

this is the answer.find the photo attached