Question

The de broglie wavelength associated with an electron is moving with a speed of 5.4×10^6m/s

Answer 1

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

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

Mass of electron (m) = ${9.1 × 10^{-31} \: Kg}$

Velocity of electron (v) = ${5.4 × 10^6 \: m/s}$

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

As we know,

$\large{\bold{ \lambda = \frac{h}{mv}}}$

Here "h" is Planck's constant which has value of $\large{6.6 \times 10^{-34 }\: m^2Kg/s}$

Substituting the values,

$\large{ \implies \lambda = \frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.4 \times 10^6}}$

$\large{ \implies \lambda = \frac{6.6 \times 10^{-34}}{9.1 \times 5.4 \times 10^{-31 + 6}}}$

$\large{ \implies \lambda = \frac{6.6 \times 10^{-34}}{49.14 \times 10^{-25}}}$

$\large{ \implies \lambda = \frac{6.6}{49.14} \times 10^{ -34 + 25}}$

$\large{ \implies \lambda = 0.13 \times 10^{-9}}$

$\huge{\boxed{\boxed{ \lambda = 1.3 \times 10^{-10} \: m}}}$

Some More Formulas:-

$\large{ \lambda = \frac{h}{p}}$

$\large{ \lambda = \frac{h}{ \sqrt{2(K.E)m}}}$

$\large{ \lambda = \frac{h}{ \sqrt{2meV}}}$

Here,

P = Linear momentum.

V = Potential difference.

e = Charge of electron.

K.E = Kinetic energy of electron.

m = mass of electron.

Answer 2

Answer:-

$\bold{\lambda = 1.25 \times 10^{-10}m} = 1.25 \times 10^-20 A\degree$

Explanation:-

Given:-

$V = 5.4 \times 10^6$

To Find:-

De-broglie wavelength( $\lambda$)

Solution:-

W.k.t,

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

${\rightarrow}\lambda = \dfrac{6.625 \times 10^-34}{9.11 \times 10^-31 \times 5.4 \times 10^6}$

${\rightarrow}\lambda = \dfrac{6.625 \times 10^{-34+31-6}}{9.11 \times 5.4 }$

${\rightarrow}\lambda = \dfrac{6.625 \times 10^{-9}}{49.194}$

${\rightarrow}\lambda = \dfrac{10^{-9}}{8}$

${\rightarrow}\lambda = 0.125 \times 10^{-9}$

${\rightarrow}\lambda = 1.25 \times 10^{-10}m$

Note:-

${\rightarrow}planck's \: constant (h)= 6.625 \times 10^{-34} \: Js$

${\rightarrow}mass \: of \: electron (m)= 9.11 \times 10^{-31} \: kg$

${\rightarrow}1\: metre= 10^{-10} A\degree$