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Question

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K.E_{max} = \frac{hc}{\lambda} -

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K.E_{max} = \frac{hc}{\lambda} - \phi$. Thus,option $A$ is correct.The stopping potential $V_0$ is related to the maximum kinetic energy by $eV_0 = K.E_{max}$. Therefore,$V_0 = \frac{1}{e}\left[ \frac{hc}{\lambda} - \phi ight]$. Thus,option $B$ is correct.The intensity of light at distance $r$ is $I = \frac{P}{4 \pi r^2}$. The energy of one photon is $E_{photon} = \frac{hc}{\lambda}$. The number of photons per unit area per unit time is $n = \frac{I}{E_{photon}} = \frac{P \lambda}{4 \pi r^2 hc}$. The number of photons striking the surface of area $A$ per unit time is $N = n 	imes A = \frac{P \lambda A}{4 \pi hc r^2}$. Thus,option $C$ is correct.Since options $A$,$B$,and $C$ are all correct statements,the wrong option is $D$.

When a light source of power $P$ emitting monochromatic light of wavelength $\lambda$ is kept at a distance $r$ from a photosensitive surface of work function $\phi$ and surface area $A$,which of the following is the wrong option?

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