(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 =

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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 $1.8 \ eV$.

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

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The surface of a certain metal is first illuminated with light of wavelength $\lambda_1 = 350 \ nm$ and then with light of wavelength $\lambda_2 = 540 \ nm$. It is found that the maximum speed of the photoelectrons in the two cases differs by a factor of $2$. The work function of the metal (in $eV$) is close to: (Energy of photon $= \frac{1240}{\lambda \text{ (in } nm)} \ eV$)

JEE MAIN 2019 All JEE MAIN

A
$1.8$
B
$2.5$
C
$5.6$
D
$1.4$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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