The work functions of two metals (MA and MB) are in the 1: 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of MA: MB is in the ratio of…

Given: The work functions satisfy A: B = 1: 2 and the photon energy is 6 eV. The kinetic energy ratio is KA: KB = 2.642: 1. Find: The work functions of MA and MB. Use Einstein's photoelectric equation: h = + K Let A =, B = 2 Then KA = 6 - KB = 6 - 2 Using the given ratio, KAKB = 6- 6-2 = 2.642 So, 6 - = 2.642(6 - 2) 6 - = 15.852 - 5.284 4.284 = 9.852 = 9.8524.284 2.3 eV Therefore, A = 2.3 eV, B = 4.6 eV The correct option is A.

The work functions of two metals (MA and MB) are in the 1: 2 rati… — Solution | JEE Chemistry

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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 $6 \, \text{eV}$ , the kinetic energy of liberated electrons of $M_A$ : $M_B$ is in the ratio of $2.642 : 1$ . The work functions (in $\text{eV}$ ) of $M_A$ and $M_B$ are respectively.

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

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The work functions satisfy $\phi_A : \phi_B = 1 : 2$ and the photon energy is $6 \, \text{eV}$ . The kinetic energy ratio is $K_A : K_B = 2.642 : 1$ .

Find: The work functions of $M_A$ and $M_B$ .

Use Einstein's photoelectric equation:

h\nu = \phi + K_{\max}

Let

\phi_A = \phi, \qquad \phi_B = 2\phi

Then

K_A = 6 - \phi

K_B = 6 - 2\phi

Using the given ratio,

\frac{K_A}{K_B} = \frac{6-\phi}{6-2\phi} = 2.642

So,

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

6 - \phi = 15.852 - 5.284\phi

4.284\phi = 9.852

\phi = \frac{9.852}{4.284} \approx 2.3 \, \text{eV}

Therefore,

\phi_A = 2.3 \, \text{eV}, \qquad \phi_B = 4.6 \, \text{eV}

The correct option is A.

Common mistakes

Taking the work function ratio $1:2$ as the kinetic energy ratio is incorrect. The given ratio of $2.642 : 1$ is for the liberated electrons' kinetic energies, so Einstein's photoelectric equation must be applied first.
Using $K = h\nu + \phi$ is wrong because the work function is the minimum energy needed to remove the electron. The correct relation is $h\nu = \phi + K_{\max}$ , so kinetic energy is obtained after subtracting the work function.
Assigning $\phi_A = 2\phi$ and $\phi_B = \phi$ reverses the given order. Since $M_A : M_B = 1 : 2$ for work functions, one must take $\phi_A = \phi$ and $\phi_B = 2\phi$ .

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