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(C) According to Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\

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$4\lambda$

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(C) According to Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.For the first case: $6eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad ...(1)$For the second case: $2eV_0 = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \quad ...(2)$Subtracting equation $(2)$ from equation $(1)$:$(6eV_0 - 2eV_0) = (\frac{hc}{\lambda} - \frac{hc}{2\lambda}) - (\frac{hc}{\lambda_0} - \frac{hc}{\lambda_0})$$4eV_0 = \frac{hc}{2\lambda} \implies eV_0 = \frac{hc}{8\lambda}$Substitute $eV_0$ into equation $(2)$:$2(\frac{hc}{8\lambda}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$$\frac{hc}{4\lambda} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$$\frac{hc}{\lambda_0} = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$Therefore,$\lambda_0 = 4\lambda$.

List-$I$	List-$II$
$(P)$ $2000 \, K$	$(1)$ The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \, eV$
$(Q)$ $3000 \, K$	$(2)$ The radiation at peak wavelength is visible to human eye.
$(R)$ $5000 \, K$	$(3)$ The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.
$(S)$ $10000 \, K$	$(4)$ The power emitted per unit area is $1/16$ of that emitted by a blackbody at temperature $6000 \, K$.
	$(5)$ The radiation at peak emission wavelength can be used to image human bones.

Similar Questions

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