The photoelectric threshold wavelength of sil | vedclass

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(A) The maximum kinetic energy is given by Einstein's photoelectric equation:$K_{\max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$where $\lambda_0 =

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$6 	imes 10^5 \, m/s$

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(A) The maximum kinetic energy is given by Einstein's photoelectric equation:$K_{\max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$where $\lambda_0 = 3250 	imes 10^{-10} \, m$ is the threshold wavelength and $\lambda = 2536 	imes 10^{-10} \, m$ is the incident wavelength.Given $hc = (4.14 	imes 10^{-15} \, eV \cdot s) 	imes (3 	imes 10^8 \, m/s) = 12420 \, eV \cdot \mathring{A}$.Converting wavelengths to $\mathring{A}$s: $\lambda_0 = 3250 \, \mathring{A}$ and $\lambda = 2536 \, \mathring{A}$.$K_{\max} = 12420 \left( \frac{1}{2536} - \frac{1}{3250} ight) \, eV = 12420 \left( \frac{3250 - 2536}{2536 	imes 3250} ight) \, eV \approx 1.076 \, eV$.Converting $K_{\max}$ to Joules: $K_{\max} = 1.076 	imes 1.6 	imes 10^{-19} \, J \approx 1.72 	imes 10^{-19} \, J$.Using $K_{\max} = \frac{1}{2}mv^2$ with $m = 9.1 	imes 10^{-31} \, kg$:$v^2 = \frac{2 	imes 1.72 	imes 10^{-19}}{9.1 	imes 10^{-31}} \approx 0.378 	imes 10^{12} \, m^2/s^2$.$v \approx 0.615 	imes 10^6 \, m/s \approx 6 	imes 10^5 \, m/s$.

The photoelectric threshold wavelength of silver is $3250 \times 10^{-10} \, m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10} \, m$ is (Given $h = 4.14 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$):

Similar Questions

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