During a photoelectric investigation, a packet of light energy, A, of 4,8 × 10⁻¹⁹ J is shone onto a metal…

Question

During a photoelectric investigation, a packet of light energy, A, of 4,8 × 10⁻¹⁹ J is shone onto a metal surface, as shown in the diagram below. The metal surface has a threshold frequency of 8,1 × 1014 Hz.

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10.1 The term for the underlined phrase "a packet of light energy" is a photon. 10.2 Threshold frequency is the minimum frequency of incident light required to eject electrons from the surface of a metal. 10.3 Step 1: Calculate the work function (φ) of the metal using the threshold frequency (f₀). The work function is given by the formula φ = h f₀, where h is Planck's constant (6.626 × 10⁻³⁴ J·s). φ = (6.626 × 10⁻³⁴ J·s) × (8.1 × 10¹⁴ Hz); φ = 5.36906 × 10⁻¹⁹ J Step 2: Compare the energy of light packet A (E_A) with the work function (φ). Given E_A = 4.8 × 10⁻¹⁹ J. Since E_A < φ (4.8 × 10⁻¹⁹ J < 5.36906 × 10⁻¹⁹ J), the energy of light A is not sufficient to eject an electron. Answer: NO 10.4 Step 1: Calculate the maximum kinetic energy (Kmax) of the ejected photoelectrons using the photoelectric effect equation. The energy of light packet B is E_B = 7.68 × 10⁻¹⁹ J. The work function φ = 5.36906 × 10⁻¹⁹ J (from 10.3). E_B = φ + Kmax; Kmax = E_B − φ; Kmax = (7.68 × 10⁻¹⁹ J) − (5.36906 × 10⁻¹⁹ J); Kmax = 2.31094 × 10⁻¹⁹ J Step 2: Calculate the maximum speed (vmax) using the maximum kinetic energy. The formula for kinetic energy is Kmax = 1/2 m_e vmax², where m_e is the mass of an electron (9.11 × 10⁻³¹ kg). vmax = √((2 Kmax)/(m_e)); vmax = √2 × (2.31094 × 10⁻¹⁹ J)9.11 × 10⁻³¹ kg; vmax = √((4.62188 × 10⁻¹⁹)/(9.11 × 10⁻³¹)); vmax = √(5.0734138 × 10¹¹); vmax = 7.12278 × 10⁵ m/s Rounding to three significant figures: vmax = 7.12 × 10⁵ m/s 10.5 The maximum speed of the ejected photoelectrons will REMAIN THE SAME. 10.6 The maximum kinetic energy (and thus maximum speed) of the ejected photoelectrons depends on the energy of individual photons (which is determined by their frequency) and the work function of the metal. The intensity of light relates to the number of photons incident per unit time, not the energy of each individual photon. Therefore, decreasing the intensity will decrease the number of ejected electrons (photoelectric current) but will not change their maximum kinetic energy or speed, as long as the frequency of the light remains constant.

During a photoelectric investigation, a packet of light energy, A, of 4,8 x 10⁻¹⁹ J is shone onto a metal surface, as shown in the diagram below. The metal surface has a threshold frequency of 8,1 x 1014 Hz.

Answer

Step-by-step solution

Part 10.1

Part 10.2

Part 10.3· 2 steps

Calculate the work function ( $\phi$ ) of the metal using the threshold frequency ( $f_0$ ).

Compare the energy of light packet A ( $E_A$ ) with the work function ( $\phi$ ).

Part 10.4· 2 steps
$7.12 \times 10^5 \text{ m/s}$

Calculate the maximum kinetic energy ( $K_{max}$ ) of the ejected photoelectrons using the photoelectric effect equation.

Calculate the maximum speed ( $v_{max}$ ) using the maximum kinetic energy.

Part 10.5

Part 10.6

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