Light of wavelength 488 , nm produced by an A | vedclass

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(A) Given: Wavelength of light, $\lambda = 488 \, nm$ and stopping potential, $V_0 = 0.38 \, V$. Energy of a photon is given by $E = \frac{hc}{\lambda

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$2.16$

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(A) Given: Wavelength of light, $\lambda = 488 \, nm$ and stopping potential, $V_0 = 0.38 \, V$. Energy of a photon is given by $E = \frac{hc}{\lambda}$. Using $hc \approx 1240 \, eV \cdot nm$ (or $1242 \, eV \cdot nm$), we calculate $E = \frac{1240}{488} \approx 2.54 \, eV$. Using $hc = 1242 \, eV \cdot nm$, $E = \frac{1242}{488} \approx 2.545 \, eV$. Maximum kinetic energy of photoelectrons is $KE_{\max} = eV_0 = 0.38 \, eV$. According to Einstein's photoelectric equation, $E = KE_{\max} + W_0$. Therefore, the work function $W_0 = E - KE_{\max} = 2.54 - 0.38 = 2.16 \, eV$. Thus, the work function of the cathode material is $2.16 \, eV$.

Light of wavelength $488 \, nm$ produced by an Argon laser is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping potential of the photoelectrons is $0.38 \, V$. The work function of the cathode material is (in $ \, eV$)

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