Question

What is the de broglie wavelength of an electron which is accelerated by 37.5 v potential difference?

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf \lambda =0.2nm}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given}$

• Potential Difference (V) = 37.5 volts

We know that :-

• Mass of electron (m) = $\sf{9.11 \times {10}^{-31}kg}$

• Charge on electron (e) = $\sf{1.6 \times {10}^{-19}C}$

$\large\underline\pink{\sf To\:Find}$

• Wavelenght $\large{\lambda}$ = ?

___________________________________

We know that ,

$\large{♡}$$\large{\boxed{\sf v=\sqrt{{\frac{2eV}{m}}}}}$

v = velocity

V = Potential Difference

On putting value

$\large\implies{\sf v=\sqrt{{\frac{2 \times 1.6 \times {10}^{-19} \times 37.5}{9.11 \times {10}^{-31}}}}}$

$\large\implies{\sf v=\sqrt{13.17 \times {10}^{12}}}$

$\large\implies{\sf v =3.6 \times {10}^{6}m/s}$

________________________________

$\large{♡}$$\large{\boxed{\sf \lambda ={\frac{h}{mv}}}}$

Here ,

h = Planck constant

its value is $\sf{6.63 \times {10}^{-34}J}$

On putting value :

$\large\implies{\sf {\frac{6.63 \times {10}^{-34}}{9.11 \times {10}^{-31} \times 3.6 \times {10}^{6}}}}$

$\large\implies{\sf {\frac{6.63 \times {10}^{-34}}{32.7 \times {10}^{-25}}}}$

$\large\implies{\sf 0.2 \times {10}^{9}}$

$\huge\red{♡}$$\large\red{\boxed{\sf \lambda =0.2nm}}$