The kinetic energy of the photoelectrons incr | vedclass

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(A) Given,the change in kinetic energy of photoelectrons is $\Delta K = 0.52 eV$. The initial wavelength is $\lambda_1 = 500 \,nm$.Using Einstein's ph

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$400$

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(A) Given,the change in kinetic energy of photoelectrons is $\Delta K = 0.52 eV$. The initial wavelength is $\lambda_1 = 500 \,nm$.Using Einstein's photoelectric equation,$K = \frac{hc}{\lambda} - \phi$.For two different wavelengths $\lambda_1$ and $\lambda_2$,the change in kinetic energy is given by:$\Delta K = K_2 - K_1 = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)$.Using $hc \approx 1242 \,eV \cdot nm$,we substitute the values:$0.52 = 1242 \left( \frac{1}{\lambda_2} - \frac{1}{500} \right)$.$\frac{0.52}{1242} = \frac{1}{\lambda_2} - 0.002$.$0.000418 = \frac{1}{\lambda_2} - 0.002$.$\frac{1}{\lambda_2} = 0.002418$.$\lambda_2 = \frac{1}{0.002418} \approx 413.5 \,nm$.Rounding to the nearest option,the new wavelength is approximately $400 \,nm$.

The kinetic energy of the photoelectrons increases by $0.52 eV$ when the wavelength of incident light is changed from $500 \,nm$ to another wavelength. The new wavelength is approximately: (in $\,nm$)

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