The threshold wavelength of tungsten is 2300 | vedclass

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by:$K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_

Vedclass · Accepted Answer

$1.5$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by:$K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} ight)$Given $hc \approx 12400 \, eV \cdot \mathring{A}$,$\lambda = 1800 \, \mathring{A}$,and $\lambda_0 = 2300 \, \mathring{A}$.Substituting the values:$K_{max} = 12400 \left( \frac{1}{1800} - \frac{1}{2300} ight)$$K_{max} = 12400 \left( \frac{2300 - 1800}{1800 	imes 2300} ight)$$K_{max} = 12400 \left( \frac{500}{4140000} ight)$$K_{max} = 12400 	imes \frac{5}{41400} \approx 1.497 \, eV$Thus,the maximum kinetic energy is approximately $1.5 \, eV$.

The threshold wavelength of tungsten is $2300 \, \mathring{A}$. If ultraviolet light of wavelength $1800 \, \mathring{A}$ is incident on it,then the maximum kinetic energy of photoelectrons would be about ............ $eV$.

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