Physics, asked by prabhavpavan9048, 1 year ago

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

Answers

Answered by Anonymous
15

\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}}

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