Question

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

Answer 1

Answer:

4.5 ×10^-20 m

Explanation:

According to the given question,

K.E. = kinetic energy = 120 eV

m = mass of electron = 9.11 × 10^ -31 kg

h = Planck's constant

= 6.63 × 10^-34 m^2.kg/s

l = wavelength

And we are asked to find the wavelength of the electron.

We know,

• l = h/root (2×m×K.E.)

Putting the values we get,

• l = h/root(2×9.11×10^(-31)×120)

• l = h/root( 2186.4 × 10^-31)

• l = h/root (218.64 ×10^-30)

• l = 6.63×10^-34/ 14.78 ×10^-15

• l = 0.45 × 10^-19

• l = 4.5 ×10^-20 m

Thus wavelength of the given electron of kinetic energy of 120 eV is 4.5 ×10^-20 m.

Answer 2

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,

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

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

Solving above,

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