The wavelength of the electron microscope is 2.472*10⁻²⁰ m.
Calculation:
Provided, accelerating voltage = 2450 V
According to de Broglie's equation:
λ = h/p
where,λ = the wavelength of the particle that moves
P= momentum of moving particle
h= plank's constant = 6.63 x 10⁻³ Js
We know that,
P = √(2mk)
P = √(2mqV)
where P= momentum of moving particle
M= mass of the moving object
K= kinetic energy of the object
V= Electric potential of the object(accelerating voltage)
q= charge of the object(electron) = e=1.6 x 10⁻¹⁹C
∴λ = h/√(2meV) = [tex]\frac{6,63 *10^{-34} }{2*9.1*10^{-31}*1.6*10^{-19}*2450 }[/tex]
=2.472*10⁻²⁰ m
S0, the wavelength of an electron microscope is 2.472*10⁻²⁰ m. Think it matches option C.
Learn more about the wavelength here:
https://brainly.com/question/13533093
#SPJ1
