The maximum kinetic energy of a photoelectron | vedclass

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(B) According to Einstein's photoelectric equation: $E = \frac{hc}{\lambda} - \phi_0$ ... $(i)$ Given that when the wavelength becomes $\lambda' = \fr

Vedclass · Accepted Answer

$\frac{hc}{3 \lambda}$

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(B) According to Einstein's photoelectric equation: $E = \frac{hc}{\lambda} - \phi_0$ ... $(i)$ Given that when the wavelength becomes $\lambda' = \frac{\lambda}{3}$,the kinetic energy becomes $E' = 4E$. Substituting these into the photoelectric equation: $4E = \frac{hc}{\lambda/3} - \phi_0$ $4E = \frac{3hc}{\lambda} - \phi_0$ ... (ii) Substitute the value of $E$ from equation $(i)$ into equation (ii): $4\left(\frac{hc}{\lambda} - \phi_0\right) = \frac{3hc}{\lambda} - \phi_0$ $\frac{4hc}{\lambda} - 4\phi_0 = \frac{3hc}{\lambda} - \phi_0$ $\frac{4hc}{\lambda} - \frac{3hc}{\lambda} = 4\phi_0 - \phi_0$ $\frac{hc}{\lambda} = 3\phi_0$ $\phi_0 = \frac{hc}{3\lambda}$

The maximum kinetic energy of a photoelectron is $E$ when the wavelength of incident radiation is $\lambda$. If the wavelength of the incident radiation is reduced to $\frac{\lambda}{3}$,the maximum kinetic energy becomes $4E$. The work function of the metal is:

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