Question

The work function of metal is \(5.3 \mathrm{eV}\). What is threshold frequency? (A) \(3.1 \times 10^{15} \mathrm{~Hz}\) (B) \(3.1 \times 10^{45} \mathrm{~Hz}\) (C) \(1.3 \times 10^{15} \mathrm{~Hz}\) (D) \(1.3 \times 10^{45} \mathrm{~Hz}\)

Short answer

The threshold frequency can be calculated using the photoelectric effect equation \(ϕ = h \times f\), where \(ϕ\) is the work function and \(h\) is Planck's constant. Converting the given work function \(5.3 \mathrm{eV}\) to Joules, we get \(8.48 \times 10^{-19} \mathrm{J}\). Using Planck's constant \(h = 6.63 \times 10^{-34} \mathrm{Js}\), we can find the threshold frequency \(f = \frac{ϕ}{h} = 1.28 \times 10^{15} \mathrm{Hz}\). Comparing with the options, the closest option is (C) \(1.3 \times 10^{15} \mathrm{Hz}\).

Step-by-step solution

Write down the photoelectric effect equation

The photoelectric effect equation relating the work function (ϕ), Planck's constant (h), and the threshold frequency (f) is: \(ϕ = h \times f\)

Identify the given work function and write it in Joules

The given work function is 5.3 eV. We need to convert it to Joules using the following relationship: 1 eV = \(1.6 \times 10^{-19} J\) So, the work function in Joules is: \(ϕ = 5.3 \mathrm{eV} \times 1.6 \times 10^{-19} \frac{\mathrm{J}}{\mathrm{eV}} = 8.48 \times 10^{-19} \mathrm{J}\)

Find Planck's constant

Planck's constant (h) has a fixed value, which is: \(h = 6.63 \times 10^{-34} \mathrm{Js}\)

Calculate the threshold frequency

Now we can use the photoelectric effect equation to find the threshold frequency: \(f = \frac{ϕ}{h}\) \(f = \frac{8.48 \times 10^{-19} \mathrm{J}}{6.63 \times 10^{-34} \mathrm{Js}}\) \(f = 1.28 \times 10^{15} \mathrm{Hz}\)

Match the calculated threshold frequency with the given options

Now that we have calculated the threshold frequency, we can compare it to the given options: (A) \(3.1 \times 10^{15} \mathrm{~Hz}\) (B) \(3.1 \times 10^{45} \mathrm{~Hz}\) (C) \(1.3 \times 10^{15} \mathrm{~Hz}\) (D) \(1.3 \times 10^{45} \mathrm{~Hz}\) The calculated threshold frequency \(1.28 \times 10^{15} \mathrm{Hz}\) is approximately equal to (C) \(1.3 \times 10^{15} \mathrm{Hz}\). Therefore, the correct answer is (C).