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

The work functions of two metals are in the 1 : 2 ratio and kinetic energy ratio question

Q. 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. 3.1, 6.2

B. 1.5, 3.0

C. 2.3, 4.6

D. 1.4, 2.8

Correct Answer: 2.3, 4.6

Explanation

According to Einstein’s photoelectric equation:

$$ \mathrm{K.E. = h\nu - \phi} $$

Given photon energy:

$$ \mathrm{h\nu = 6\ eV} $$

Let the work function of metal M_A be $\phi_A = x$ eV.

Since the work functions are in the ratio 1 : 2, the work function of metal M_B is:

$$ \mathrm{\phi_B = 2x} $$

Kinetic energy of electrons from M_A:

$$ \mathrm{K_A = 6 - x} $$

Kinetic energy of electrons from M_B:

$$ \mathrm{K_B = 6 - 2x} $$

Given ratio of kinetic energies:

$$ \mathrm{\frac{K_A}{K_B} = 2.642} $$

Substituting values:

$$ \mathrm{\frac{6 - x}{6 - 2x} = 2.642} $$

Cross multiplying:

$$ \mathrm{6 - x = 2.642(6 - 2x)} $$

$$ \mathrm{6 - x = 15.852 - 5.284x} $$

$$ \mathrm{5.284x - x = 15.852 - 6} $$

$$ \mathrm{4.284x = 9.852} $$

$$ \mathrm{x = 2.3\ eV} $$

Therefore:

$$ \mathrm{\phi_A = 2.3\ eV,\ \phi_B = 4.6\ eV} $$

Hence, the correct answer is 2.3 eV and 4.6 eV.

Explanation

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