Question

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of 9.76×106 m/s .What is the characteristic wavelength of this electron? (Mass of electron is 9.1094×10−31 kg)

Answer 1

Use the de Broglie relationship between mass and wavelength (lambda).

lambda=h(Planck's constant)/[(mass)*(volume)]

lambda=6.626*10^-34/[(9.76*10^6)*(9.1094*10^-31)]

=7.45*10^-11 m

Answer 2

Answer is 7.45 x 10⁻¹¹ m

Explanation;

To solve this problem we can use De Broglie equation,

∧ = h / mv

Where, ∧ is the wave length (m), h is the Planck's constant (6.626 × 10⁻³⁴ m² kg / s), m is the mass (kg) and v is the velocity of the object (m/s).

∧ = ?

h = 6.626 × 10⁻³⁴ m² kg / s

v = 9.76 × 10⁶ m/s

m = 9.1094 × 10⁻³¹ kg

From substitution,

∧ = (6.626 × 10⁻³⁴ m² kg / s ) / ( 9.76 × 10⁶ m/s x 9.1094 × 10⁻³¹ kg)

∧ = 7.45 x 10⁻¹¹ m