Question

the mass of an electron is 9.1 × 10^-31 kg. if its k.e is 3.0 × 10^-25 J, calculate its wavelength. structure of atom.​

Answer 1

Answer:

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Explanation:

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Answer 2

Given :-

• Mass of electron = 9.1 × 10⁻³¹ kg.

• Kinetic energy of electron = 3.0 × 10⁻²⁵ J

To Find :-

• Wavelength (λ) of the electron.

Solution :-

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

• $\rm{K.E=\dfrac{1}{2}mv^2}$

• $\rm{v=\left(\dfrac{2K.E}{m}\right)^{1/2}}$

$\underline{\:\textsf{ Putting the given values :}}$

$\dashrightarrow \rm{v=\left(\dfrac{2\times{}3.0\times{}10^{-25}\:kgm^2s^{-2}}{9.1\times{}10^{-31}\:kg}\right)^{1/2}}$

$\dashrightarrow \rm{v=812\:ms^{-1}}$

$\underline{\:\textsf{Using Brogile equation :}}$

$\rm{\lambda=\dfrac{h}{mv}}$

$\sf Where$

•Planck's constant, h = 6.626 × 10⁻³⁴ Js.

•Mass of electron, m = 9.1 × 10⁻³¹ kg.

$\underline{\:\textsf{ Putting the given values :}}$

$\dashrightarrow \rm{\lambda=\dfrac{6.626\times{}10^{-34}\:kgm^2s^{-1}}{(9.1\times{}10^{-31}\:kg)(812\:ms^{-1})}}$

$\dashrightarrow \rm{\lambda=8967\times{}10^{-10}\:m}\:$

$\dashrightarrow {\boxed{\frak{\purple{\lambda= 896.7\;nm}}}}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Wavelength</p><p>\textbf{ 896.7 nm }}}.$

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