Question

Q1.What is the de Broglie wavelength
of an electron whose k.E. is 120ev?
​

Answer 1

Step-by-step explanation:

Answer:

$\lambda=1.12\;\;A\textdegree$

Explanation:

Given,

Kinetic Energy = 120 eV

We know that,

1 eV = 1.6 × 10⁻¹⁹ Joule

∴ 120eV = 120 × 1.6 × 10⁻¹⁹ Joule

= 1.92 × 10⁻¹⁷ Joule.

If the mass of electron is m(=9.1×10⁻31 kg) and velocity is v,

K.E. = 1/2mv²

1.92 × 10⁻¹⁷ = 1/2 × 9.1 ×10⁻³¹ × v²

3.84 × 10⁻¹⁷ = 9.1 × 10⁻³¹ × v²

v² = 42.198 × 10¹²

v = 6.495 × 10⁶ m/s

Using de - Broglie's formula,

$\begin{gathered}\lambda=\frac{h}{mv}\\\;\\\lambda=\frac{6.625\times10^{-34}}{9.1\times10^{-31}\times6.495\times10^{6}}\end{gathered}$

Here h = 6.625 × 10⁻³⁴ Joule-Second which is Plank's constant.

Solving above,

$\begin{gathered}\lambda=1.12\times10^{-10}\;\;m\\\;\\\lambda=1.12\;\;A\textdegree\end{gathered}$