Question

The mass of an electron is 9.1 × $10^{-28}$ g. If its K.E. is 3.0 × $10^{-25}$ J, calculate its wavelength in Angstrom.

Answer 1

"A particle's de-Broglie wavelength is defined as Planck's constant divided by the particle's momentum

Kinetic energy of electron $=\quad \frac { 1 }{ 2 } m{ v }^{ 2 }$

$=\sqrt { \frac { 2\quad \times \quad 3.0\quad \times \quad { 10 }^{ -25 }\quad J }{ 9.11\quad \times \quad { 10 }^{ -31 }\quad Kg}}$

From the above

$v\quad =\quad \frac { \sqrt { 2 } \quad \times \quad K.E }{ m } \quad =\quad 812\quad m{ s }^{ -1 }$

By de-Broglie equation,

$\lambda \quad =\quad \frac { h }{ mv } \quad =\quad \frac { 6.626\quad \times \quad { 10 }^{ -34 }\quad Js }{ (9.11\quad \times \quad { 10 }^{ -31 }\quad kg)(812) }$

$\lambda \quad =\quad 8.967\quad \times \quad { 10 }^{ -7 }\quad m$

Therefore, wavelength of electron is $8.967\quad \times \quad { 10 }^{ -7 }\quad m$"

Answer 2

Answer:

=h/(2mK.E)

1/2

λ=6.626×10

−34

/(2×9.1×10

−31

×3×10

−25

)

1/2

=1.2×10

−7

m