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(A) Let the work function be $\phi$. The energy of the incident photons is given by $E = \frac{1240}{\lambda} \ eV$.For $\lambda_1 = 350 \ nm$,$E_1 =

Vedclass · Accepted Answer

$1.8$

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(A) Let the work function be $\phi$. The energy of the incident photons is given by $E = \frac{1240}{\lambda} \ eV$.For $\lambda_1 = 350 \ nm$,$E_1 = \frac{1240}{350} \approx 3.543 \ eV$.For $\lambda_2 = 540 \ nm$,$E_2 = \frac{1240}{540} \approx 2.296 \ eV$.According to Einstein's photoelectric equation,$KE_{max} = E - \phi$.Let $v_1$ and $v_2$ be the maximum speeds. Given $v_1 = 2v_2$,so $KE_1 = 4 KE_2$.$3.543 - \phi = 4(2.296 - \phi)$.$3.543 - \phi = 9.184 - 4\phi$.$3\phi = 9.184 - 3.543 = 5.641$.$\phi = \frac{5.641}{3} \approx 1.88 \ eV$.The closest value is $1.8 \ eV$.

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