In a technique used for surface analysis called auger electron spectroscopy (AES), electrons are accelerated toward a metal surface. These electrons cause the emissions of secondary electrons (called auger electrons) from the metal surface. The kinetic energy of the auger electrons depends on the composition of the surface. The presence of oxygen atoms on the surface results in auger electrons with a kinetic energy of approximately 520 eV . Required:
What is the de Broglie wavelength of one of these electrons?

the de Broglie wavelength of one of the electrons is 5.3787 × 10⁻¹¹ m

Explanation:

Given the data in the question;

we know that; ( electron volt ) 1 eV = 1.602 × 10⁻¹⁹ J

so given that Kinetic Energy KE = 520 eV

KE = 520 × 1.602 × 10⁻¹⁹ = 8.3304 × 10⁻¹⁷ J

we know that;

Kinetic Energy = 1/2 × mv²

where m is the electron mass

and mass of electron is  9.109 × 10⁻³¹ kilograms

so

8.3304 × 10⁻¹⁷ = 1/2 × 9.109 × 10⁻¹⁷× v²

v = √[(8.3304 × 10⁻¹⁷) / (1/2 × 9.109 × 10⁻³¹)]

v = 1.3524 × 10⁷ m/s

speed of the electron v = 1.3524 × 10⁷ m/s

now, de Broglie equation;

wavelength λ = h / mv

where h is Planck constant ( 6.626 × 10⁻³⁴ m² kg / s )

so we substitute

wavelength λ = 6.626 × 10⁻³⁴ / ( 9.109 × 10⁻³¹ × 1.3524 × 10⁷)

wavelength λ = 5.3787 × 10⁻¹¹ m

Therefore, the de Broglie wavelength of one of the electrons is 5.3787 × 10⁻¹¹ m

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answer: they produce more slow than unicellular organisms.