The work functions of two metals (M_A and M_B | vedclass

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(B) According to Einstein's photoelectric equation, $KE_{max} = E - \phi$. Given that the ratio of work functions $\phi_A : \phi_B = 1:2$, let $\phi_A

Vedclass · Accepted Answer

$2.3, 4.6$

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(B) According to Einstein's photoelectric equation, $KE_{max} = E - \phi$. Given that the ratio of work functions $\phi_A : \phi_B = 1:2$, let $\phi_A = \phi$ and $\phi_B = 2\phi$. For metal $M_A$: $(KE_{max})_A = 6 - \phi$. For metal $M_B$: $(KE_{max})_B = 6 - 2\phi$. The ratio of kinetic energies is given as $\frac{(KE_{max})_A}{(KE_{max})_B} = \frac{2.642}{1}$. Substituting the expressions: $\frac{6 - \phi}{6 - 2\phi} = 2.642$. $6 - \phi = 2.642(6 - 2\phi)$. $6 - \phi = 15.852 - 5.284\phi$. $4.284\phi = 9.852$. $\phi \approx 2.3 \ eV$. Thus, $\phi_A = 2.3 \ eV$ and $\phi_B = 4.6 \ eV$.

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.

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