Question

A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 1.35 nm apart. You observe that the first double-slit interference dark fringe occurs at ±15.0 ∘ from the original direction of the beam when viewed on a distant screen. Through what potential difference were the electrons accelerated?

Answer 1

Answer:

The potential difference was 3.09V.

Explanation:

The condition for the first dark fringe is

$(1).\: \: \: d\:sin(\theta) = \dfrac{\lambda}{2}$

$\lambda = 2d\: sin(\theta)$.

putting in numbers we get:

$\lambda = 2 (1.35*10^{-9})(sin(15^o))$

$\boxed{\lambda = 6.988*10^{-10}m}$

Now, from de broglie relation we know that

$\lambda = \dfrac{h}{p}$

or

$(2). \: \: p = \dfrac{h}{\lambda }$

where $h$  is Planck's constant and $p$ is electron momentum. The momentum is related to kinetic energy by

$E = \dfrac{p^2}{2m},$

this energy we know must come from potential difference; therefore,

$(3). \: \: qV= \dfrac{p^2}{2m}$

substituting the value of $p$ from equation (2), and solving for potential $V$, we get:

$(4). \: \:V = \dfrac{h^2}{2qm\lambda^2}$

putting in numerical values in equation (4) we get:

$V = \dfrac{(6.63*10^{-34})^2}{2(1.6*10^{-19})(9.1*10^{-31})(6.988*10^{-10})^2}$

$\boxed{V = 3.09\:volts}$.

Therefore, the potential difference through which the protons were accelerated was 3.09V.