Question

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6E3 km*s^-1. (a) What is the wavelength of the ejected electron? (b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50E16 Hz. How much energy is required to remove the electron from the metal surface? (c) What is the wavelength of the radiation that caused photoejection of the electron? (d) What kind of electromagnetic radiation was used?

Answer 1

(a) The wavelength of the electron is 202.25885 nm

(b) The minimum energy required to remove the electron is 1.6565 × 10⁻¹⁷ J

(c) The wavelength of the causing radiation is approximately 8.84 nm

(d) X-ray

The question parameters are;

The given parameters of the electron are;

The velocity of the electron, v = 3.6 × 10³ km/s

(a) de Broglie wavelength is given as follows;

λ = h/(m·v)

Where;

λ = The wavelength of the wave

h = Planck's constant = 6.626 × 10⁻³⁴ J·s

m = The mass of the electron = 9.1 × 10⁻³¹ kg

Therefore, we get;

λ = 6.626 × 10⁻³⁴/(9.1 × 10⁻³¹ × 3.6 × 10⁶) = 202.25885 × 10⁻⁶

The wavelength, λ, of the electron is 202.25885 × 10⁻⁶ m = 202.25885 nm

(b) The energy required to remove the electron from the metal surface is known as the work function, W₀, which is given by the following formula

W₀ = h·f₀

Where;

f₀ = The threshold frequency

Given that the threshold frequency, f₀ = 2.50 × 10¹⁶ Hz, we have;

W₀ = 6.626 × 10⁻³⁴ J·s × 2.50 × 10¹⁶ Hz = 1.6565 × 10⁻¹⁷ J

The energy required to remove the electron from the metal surface, W₀ = 1.6565 × 10⁻¹⁷ J

(c) The wavelength of the radiation that caused the photoejection of the electron is given as follows;

The energy of the incoming photon, E = W₀ + (1/2)·m·v²

Where;

v = The velocity of the electron, and m = The mass of the electron

Therefore;

E = 1.6565 × 10⁻¹⁷ + (1/2) × 9.1 × 10⁻³¹ kg × (3.6 × 10⁶ m/s)² = 2.24618 × 10⁻¹⁷ J

We have;

E = h·f

∴ f = (2.24618 × 10⁻¹⁷ J)/(6.626 × 10⁻³⁴ J·s) = 3.38994869 × 10¹⁶ Hz

The speed of light, c = 299,792,458 m/s

From the equation for the speed of light, we have;

λ = c/f

∴ λ = (299,792,458 m/s)/(3.38994869 × 10¹⁶ Hz) = 8.84356919 nm ≈ 8.84 nm

The wavelength of the radiation that caused photoejection of the electron, λ$_{causing \ radiation}$ ≈ 8.84 nm

(d) The kind of electromagnetic radiation used which has a wavelength of 8.84 nm is the X-Ray which are electromagnetic radiation having wavelengths that extend from 10 picometers to 10 nanometers.

Learn more about De Broglie wavelength here;

brainly.com/question/19131384