(A) The energy of the incident photons is given by $E = \frac{hc}{\lambda}$.For $\lambda_1 = 248 \ nm$,$E_1 = \frac{1240}{248} = 5 \ eV$.For $\lambda_

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(A) The energy of the incident photons is given by $E = \frac{hc}{\lambda}$.For $\lambda_1 = 248 \ nm$,$E_1 = \frac{1240}{248} = 5 \ eV$.For $\lambda_2 = 310 \ nm$,$E_2 = \frac{1240}{310} = 4 \ eV$.According to Einstein's photoelectric equation,the maximum kinetic energy is $K.E. = E - W$,where $W$ is the work function.Thus,$K.E._1 = 5 - W$ and $K.E._2 = 4 - W$.Since $K.E. = \frac{1}{2}mv^2$,the ratio of kinetic energies is $\frac{K.E._1}{K.E._2} = \left(\frac{u_1}{u_2}\right)^2 = \left(\frac{2}{1}\right)^2 = 4$.Therefore,$\frac{5 - W}{4 - W} = 4$.$5 - W = 4(4 - W) = 16 - 4W$.$3W = 11$.$W = \frac{11}{3} \approx 3.67 \ eV \approx 3.7 \ eV$.

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

$A$ metal surface is illuminated by light of two different wavelengths $248 \ nm$ and $310 \ nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$,respectively. If the ratio $u_1: u_2 = 2: 1$ and $hc = 1240 \ eV \ nm$,the work function of the metal is nearly: (in $eV$)

IIT 2014 All IIT

A
$3.7$
B
$3.2$
C
$2.8$
D
$2.5$

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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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When a photon of wavelength ${\lambda _1}$ is incident on a metal surface,the speed of the ejected fastest electron is $v_1 \ m/s$. If a photon of wavelength ${\lambda _2}$ is incident on the same metal,the speed of the ejected fastest electron is $v_2 \ m/s$. The value of $v_2^2 - v_1^2$ will be

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