Question

On two metal surfaces, a monochromatic light of $6 \text{ eV}$ was incident. They have ratio of their work function and maximum KE as $\frac{\Phi_1}{\Phi_2} = \frac{1}{2}$ and $\frac{(KE_{\max})_1}{(KE_{\max})_2} = \frac{2.62}{1}$. Then $\Phi_1$ and $\Phi_2$ values are respectively (in $\text{eV}$).

• A. $2.292, 4.584$
• B. $4.584, 2.292$
• C. $4.584, 9.168$
• D. $1.146, 2.292$

Hint:

When given ratios of work functions and kinetic energies, express both KE terms in relation to the work functions using $KE = E - \Phi$, and solve the resulting system of simultaneous equations.

Answer 1

The Correct Option is A

To solve this problem, we apply the photoelectric effect equation, which is given as:

\(E = \Phi + KE_{\max}\) 

where:

• \(E\) is the energy of the incident photon.
• \(\Phi\) is the work function of the metal.
• \(KE_{\max}\) is the maximum kinetic energy of the emitted electron.

We are given:

• Photon energy \(E = 6 \text{ eV}\)
• The ratio of work functions: \(\frac{\Phi_1}{\Phi_2} = \frac{1}{2}\)
• The ratio of maximum kinetic energies: 6 = \Phi_1 + (KE_{\max})_1 (Equation 1)

 

\(6 = \Phi_2 + (KE_{\max})_2\) (Equation 2)

From the given ratios, we can express:

\(\Phi_1 = \frac{1}{2} \Phi_2 \quad (Equation 3)\)

\((KE_{\max})_1 = 2.62 \times (KE_{\max})_2 \quad (Equation 4)\)

Now, substitute for \(\Phi_1\) and 6 = \frac{1}{2} \Phi_2 + 2.62 \times (KE_{\max})_2

Equation 2 can be rewritten as:

into Equation 1:

\(6 = \frac{1}{2} \Phi_2 + 2.62 \times (6 - \Phi_2)\)

Simplifying the equation:

6 = 15.72 - 2.12 \Phi_2

2.12 \Phi_2 = 9.72

 

Substituting the value of

and