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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work

Vedclass · Accepted Answer

$K_{1} < \frac{K_{2}}{3}$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the metal.For wavelength $\lambda_{1}$,$K_{1} = \frac{hc}{\lambda_{1}} - \phi$.For wavelength $\lambda_{2}$,$K_{2} = \frac{hc}{\lambda_{2}} - \phi$.Given $\lambda_{1} = 3 \lambda_{2}$,we substitute this into the equation for $K_{1}$:$K_{1} = \frac{hc}{3 \lambda_{2}} - \phi$.Multiplying by $3$,we get $3 K_{1} = \frac{hc}{\lambda_{2}} - 3 \phi$.Since $K_{2} = \frac{hc}{\lambda_{2}} - \phi$,we have $\frac{hc}{\lambda_{2}} = K_{2} + \phi$.Substituting this into the expression for $3 K_{1}$:$3 K_{1} = (K_{2} + \phi) - 3 \phi = K_{2} - 2 \phi$.Since the work function $\phi > 0$,it follows that $3 K_{1} < K_{2}$,which implies $K_{1} < \frac{K_{2}}{3}$.

Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$,respectively,are incident on a metallic surface. If $\lambda_{1} = 3 \lambda_{2}$,then:

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