Question

An electron has a mass 9.108x10^-31 kg and
velocity 0.6x10^8 m/s.Calculate the
wavelength of the electron.​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of electron = 9.1 × 10^-31 Kg.
• Velocity of electron (v) = 0.6 × 10⁸ m/s.

$\huge{\bold{\underline{Explanation:-}}}$

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$\large{\boxed{ \tt \lambda = \dfrac{h}{mv}}}$

Substituting the values,

$\large{\tt \lambda = \dfrac{6.626 \times 10^{-34}}{ 9.1 \times 10^{-31} \times 0.6 \times 10^8}}$

Note:-

$\large{\tt h = Planck's \: constant}$

Now,

$\large{\tt \lambda = \dfrac{6.626 \times 10^{-34}}{9.1 \times 0.6 \times 10^{-31+8}}}$

$\large{\tt \lambda = \dfrac{6.626 \times 10^{-34}}{5.46 \times 10^{-23}}}$

It comes as,

$\large{\tt \lambda = \dfrac{6.626 \times 10^{-34 + 23}}{5.46}}$

$\large{ \tt \lambda = \dfrac{6.626 \times 10^{-11}}{5.46}}$

$\huge{\boxed{\boxed{\tt \lambda = 1.21 \times 10^{-11} \: meters }}}$

So, the wavelength of electron is 1.21 × 10^{-11} meters .

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Additional formulas:-

• $\large{\tt \lambda = \dfrac{h}{p}}$
• $\large{\tt \lambda = \dfrac{h}{ \sqrt{2(K.E)m}}}$
• $\large{\tt \lambda = \dfrac{h}{ \sqrt{2meV}}}$

Note:-

• $\large{\tt p = momentum}$
• $\large{\tt m = Mass \: of \: electron}$
• $\large{\tt V = Potential \: difference}$
• $\large{\tt e = charge \: of \: electron}$
• $\large{\tt h = Planck's \: constant}$

Values:-

• $\large{\tt e = 1.6 \times 10^{-19} C}$
• $\large{\tt h = 6.26 \times 10^{-34} \: Js}$
• $\large{\tt m= 9.108 \times 10^{-31} \: Kg}$

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