When drawing a covalent lewis dot diagram its important to ensure that:
1 answer:
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(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 <em>m</em> = 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 ≈ 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;