Question

what is the potential difference through which an electron with the de Broglie wavelength of 1.5 Angstrom should be accelerated if its de Broglie wavelength should be reduced to 1 angstrom

Answer 1

Answer:

Potential difference V=601.98ν

Explanation:

Here OP means 0.5 Angstrom

so

λ=$0.5*10^{-10} m$

it becomes

p=$\frac{6.62*10^{-34} }{0.5*10^{-10} }$

p=$13.24*10^{-24} Ns$

mv=$13.24*10^{-24}$

v=$\frac{13.24*10^{-24} }{9.1*10^{-31} }$

v=$1.4549*10^{7} m$

E=qV

$\frac{p^{2} }{2m} =q*V$

$\frac{(13.24*10^{-24})^{2}  }{2*9.1*10^{-31} } =1.6*10^{-19} *V$

V=601.98v