The energy of a photon with wavelength lambda | vedclass

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(D) According to Einstein's photoelectric equation: $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$. Given $\frac{hc}{\lambda} = 2 \ eV$,so $\frac{1}{2}

Vedclass · Accepted Answer

$1.8$

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(D) According to Einstein's photoelectric equation: $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$. Given $\frac{hc}{\lambda} = 2 \ eV$,so $\frac{1}{2}mv^2 = 2 - \phi$  ---$(1)$ When $\lambda' = \lambda - 0.25\lambda = 0.75\lambda = \frac{3}{4}\lambda$ and $v' = 2v$,the new equation is: $\frac{1}{2}m(2v)^2 = \frac{hc}{0.75\lambda} - \phi$ $4(\frac{1}{2}mv^2) = \frac{4}{3}(\frac{hc}{\lambda}) - \phi$ Substituting $\frac{1}{2}mv^2 = 2 - \phi$ and $\frac{hc}{\lambda} = 2$: $4(2 - \phi) = \frac{4}{3}(2) - \phi$ $8 - 4\phi = \frac{8}{3} - \phi$ $8 - \frac{8}{3} = 3\phi$ $\frac{24 - 8}{3} = 3\phi$ $\frac{16}{3} = 3\phi$ $\phi = \frac{16}{9} \approx 1.777 \ eV \approx 1.8 \ eV$.

The energy of a photon with wavelength $\lambda$ is $2 \ eV$. When it strikes a metal surface,the maximum velocity of the emitted photoelectrons is $v$. If the value of $\lambda$ is decreased by $25\%$ and the maximum velocity is doubled,the work function of the metal becomes ...... $eV$.

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