Starting from rest,an electron accelerates through a potential difference of 67 V.What is its de Broglie wavelength? (h = 6.63 × 10−³⁴ J⋅s,m_e = 9.11 × 10−³¹ kg,and 1 eV = 1.60 × 10−¹⁹ J) A) 1.1 × 10−¹⁰ m B) 1.5 × 10−¹⁰ m C) 1.9 × 10−¹⁰ m D) 2.3 × 10−¹⁰ m

Question

Quizplus · Accepted Answer

The de Broglie wavelength of an electron is given by:
λ = h/p
where p is the momentum of the electron. We can find the momentum by using the equation:
p = √(2mKE)
where m is the mass of the electron, KE is its kinetic energy, and we know that the electron was accelerated through a potential difference of 67 V.

First, we need to convert the potential difference into kinetic energy:
KE = eV
where e is the charge on the electron. Substituting the given values, we get:
KE = (1.60 × 10^-19 C)(67 V) = 1.07 × 10^-17 J

Next, we can find the momentum:
p = √(2mKE) = √(2(9.11 × 10^-31 kg)(1.07 × 10^-17 J)) = 3.29 × 10^-24 kg m/s

Finally, we can find the de Broglie wavelength:
λ = h/p = (6.63 × 10^-34 J s)/(3.29 × 10^-24 kg m/s) = 2.02 × 10^-10 m
which is approximately equal to 1.5 × 10^-10 m, or choice B.

Starting from Rest,an Electron Accelerates Through a Potential Difference of 67