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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)

Respuesta :

Ishankahps

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


onyebuchinnaji

The characteristic wavelength of the given electron is 7.452 x 10⁻¹¹ m.

The given parameters;

  • speed of the electron, v = 9.76 x 10⁶ m/s
  • mass of the electron, m = 9.11 x 10⁻³¹ kg

The characteristic wavelength of the given electron is calculated by applying De Broglie equation as follows;

[tex]\lambda = \frac{h}{p} = \frac{h}{mv} \\\\\lambda = \frac{(6.626 \times 10^{-34})}{(9.11\times 10^{-31})\times (9.76\times 10^6)} \\\\\lambda = 7.452 \times 10^{-11} \ m[/tex]

Thus, the characteristic wavelength of the given electron is 7.452 x 10⁻¹¹ m.

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