Assertion : Photosensitivity of a metal is hi | vedclass

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(A) The photosensitivity of a metal refers to its ability to emit photoelectrons when light of a suitable frequency is incident on it. A metal with a

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If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

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(A) The photosensitivity of a metal refers to its ability to emit photoelectrons when light of a suitable frequency is incident on it. A metal with a low work function $(\Phi_0)$ requires less energy to eject an electron, making it more photosensitive. Thus, the Assertion is correct.The work function is defined as $\Phi_0 = hf_0$, where $h$ is Planck's constant and $f_0$ is the threshold frequency. This is a standard physical relationship for the photoelectric effect. Thus, the Reason is also correct.Since the work function is defined by the threshold frequency $(f_0)$, and a smaller $f_0$ (and thus smaller $\Phi_0$) makes it easier to eject electrons, the Reason correctly explains why a small work function leads to high photosensitivity. Therefore, the correct option is $A$.

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.

Assertion : Photosensitivity of a metal is high if its work function is small.
Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.

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Assertion : Photosensitivity of a metal is high if its work function is small.Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.