Question

The work functions of two metals ($M_A$ and $M_B$) are in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of $M_A$ : $M_B$ is in the ratio of 2.642 : 1. The work functions (in eV) of $M_A$ and $M_B$ are respectively.

• A. 2.3, 4.6
• B. 3.1, 6.2
• C. 1.4, 2.8
• D. 1.5, 3.0

Hint:

Einstein's Photoelectric Equation: $h\nu = \phi + K_{max}$.

Answer 1

The Correct Option is A

To solve this problem, we need to understand the photoelectric effect, where the kinetic energy of ejected electrons can be determined using the formula:

\(K.E. = E_{photon} - \phi\)

where \(E_{photon}\) is the photon energy, and \(\phi\) is the work function of the metal.

Let's denote the work functions of metals \(M_A\) and \(M_B\) as \(\phi_A\) and \(\phi_B\) respectively. According to the problem:

• The ratio of the work functions is \(\phi_A : \phi_B = 1 : 2\).
• Photon energy \(E_{photon} = 6 \, \text{eV}\).
• The kinetic energy of the liberated electrons is in the ratio K.E._A = 6 \, \text{eV} - \phi_A

 

\(K.E._B = 6 \, \text{eV} - \phi_B\)

Since \(\phi_B = 2\phi_A\), we can substitute and compare the kinetic energy equations:

\(\frac{6 - \phi_A}{6 - 2\phi_A} = 2.642\)

Solving this equation for \(\phi_A\):

\(6 - \phi_A = 2.642(6 - 2\phi_A)\)

\(6 - \phi_A = 15.852 - 5.284\phi_A\)

\(4.284\phi_A = 9.852\)

\(\phi_A = \frac{9.852}{4.284} \approx 2.3 \, \text{eV}\)

Since \(\phi_B = 2\phi_A\), we have:

\(\phi_B = 2 \times 2.3 = 4.6 \, \text{eV}\) 

Therefore, the work functions of \(M_A\) and \(M_B\) are 2.3 eV and 4.6 eV respectively, which matches the correct answer option.