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(B) According to Einstein's photoelectric equation,the kinetic energy $E$ is given by:$E = \frac{hc}{\lambda} - \phi$ --- $(i)$When the kinetic energy

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$\frac{ hc \lambda}{ E \lambda+ hc }$

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(B) According to Einstein's photoelectric equation,the kinetic energy $E$ is given by:$E = \frac{hc}{\lambda} - \phi$ --- $(i)$When the kinetic energy is doubled to $2E$,let the new wavelength be $\lambda^{\prime}$:$2E = \frac{hc}{\lambda^{\prime}} - \phi$ --- (ii)Subtracting equation $(i)$ from equation (ii):$(2E - E) = (\frac{hc}{\lambda^{\prime}} - \phi) - (\frac{hc}{\lambda} - \phi)$$E = \frac{hc}{\lambda^{\prime}} - \frac{hc}{\lambda}$Rearranging the terms to solve for $\lambda^{\prime}$:$E + \frac{hc}{\lambda} = \frac{hc}{\lambda^{\prime}}$$\frac{E\lambda + hc}{\lambda} = \frac{hc}{\lambda^{\prime}}$$\lambda^{\prime} = \frac{hc\lambda}{E\lambda + hc}$

The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of:

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