The work functions of two metals ( M A ? a n d 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

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: [2026]

Smart Achievers · Accepted Answer

(1)

$K E_{\max} = E - ϕ$

${(K E_{\max})}_{1} = 6 - ϕ_{1} . . . . (1)$

${(K E_{\max})}_{2} = 6 - ϕ_{2} . . . . (2)$

$By eq. (1) divide eq. (2)$

$\frac{{(K E_{\max})}_{1}}{{(K E_{\max})}_{2}} = \frac{2.642}{1} = \frac{6 - ϕ_{1}}{6 - ϕ_{2}}$

$\frac{2.642}{1} = \frac{6 - ϕ_{1}}{6 - 2 ϕ_{1}}$

$ϕ_{1} = 2.3 eV$

$ϕ_{2} = 4.6 eV$

Solution

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: [2026]

1 2.3, 4.6

2 3.1, 6.2

3 1.5, 3.0

4 1.4, 2.8

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