Question

If ratio of threshold frequencies of two metals is \(1: 3\), ratio of their work functions is \(\ldots \ldots\) (A) \(1: 3\) (B) \(3: 1\) (C) \(4: 16\) (D) \(16: 4\)

Short answer

The ratio of the work functions of the two metals is \(1:3\), which corresponds to the answer (A).

Step-by-step solution

Write the photoelectric effect equation for both metals

The photoelectric effect equation is given by: \(E = h \nu = \phi + KE\) where: - \(E\) is the energy of the incident photon - \(h\) is the Planck's constant - \(\nu\) is the frequency of the incident photon - \(\phi\) is the work function of the metal - \(KE\) is the kinetic energy of the ejected electron Now let's write the photoelectric effect equation for both the metals: For metal 1: \(E_1 = h \nu_1 = \phi_1 + KE_1\) For metal 2: \(E_2 = h \nu_2 = \phi_2 + KE_2\)

Use the given ratio of their threshold frequencies

The given ratio of threshold frequencies of the two metals is 1:3. Therefore, we can write it as: \(\frac{\nu_1}{\nu_2} = \frac{1}{3}\) Now let's substitute this ratio in the photoelectric effect equations to find the work functions: \(E_1 = h \cdot \frac{1}{3} \nu_2 = \phi_1 + KE_1\) \(E_2 = h \nu_2 = \phi_2 + KE_2\)

Determine the ratio of their work functions

Now we need to find the ratio of their work functions, which can be written as: \(\frac{\phi_1}{\phi_2}\) Since \(E_1 = \frac{1}{3} E_2\), we can write the equations as: \(\phi_1 = \frac{1}{3} \phi_2\) \(\frac{\phi_1}{\phi_2} = \frac{\frac{1}{3} \phi_2}{\phi_2}\) Simplifying the above equation, we get: \(\frac{\phi_1}{\phi_2} = \frac{1}{3}\)

Identify the correct answer

Now we can see that the ratio of the work functions of the two metals is \(1:3\), which corresponds to the answer (A).